Introductory Lectures on Quantum Field Theory

Daniele Dominici

Department of Physics and Astronomy, University of Florence, and Sezionedi Firenze, INFN, Via Sansone 1, 50019 Sesto F. (FI), Italy

Contents

1 Vibrational modes and one dimensional lattice 11.1 Lagrangian and Hamiltonian . . . . . . . . . . . . . . . . . . . 11.2 Expansion in eigenmodes and quantization . . . . . . . . . . . 41.3 Continuum limit . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Lagrangian field theory 102.1 Action and Euler equations for a continuous system . . . . . . 10

3 Quantization of the Klein-Gordon field 113.1 Quantization of the Klein-Gordon field in 1D . . . . . . . . . . 113.2 Quantization of Klein-Gordon field in 3D . . . . . . . . . . . . 143.3 Noether Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 16

4 Quantization of the electromagnetic field 224.1 Lagrangian and Hamiltonian . . . . . . . . . . . . . . . . . . . 224.2 Casimir effect . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

5 Hamiltonian for a system of non relativistic charged particlesinteracting through the electromagnetic field 30

6 Scattering theory 336.1 S matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336.2 Fermi golden rule . . . . . . . . . . . . . . . . . . . . . . . . . 40

7 Radiation processes of first order: emission and absorptionof a single photon 41

8 Interaction of the light with the matter 44

9 Cherenkov effect 51

10 The Dirac field 5510.1 The Dirac equation: classical theory . . . . . . . . . . . . . . . 5510.2 Lorentz and parity transformation . . . . . . . . . . . . . . . . 5610.3 Wave plane solutions of the Dirac equation . . . . . . . . . . . 6010.4 Lagrangian of the Dirac field . . . . . . . . . . . . . . . . . . . 6210.5 Non relativistic limit of the Dirac equation . . . . . . . . . . . 63

10.6 Quantization of the Dirac field . . . . . . . . . . . . . . . . . . 65

11 Higgs decay width to fermions 68

12 The decay width for the process µ− → e−νeνµ 73

13 Superfluidity 8013.1 Brief introduction to superfluids . . . . . . . . . . . . . . . . . 8113.2 Bose Einstein condensation for an ideal gas . . . . . . . . . . . 8513.3 Quantization of the Schrodinger field . . . . . . . . . . . . . . 8813.4 Ginzburg-Landau Model . . . . . . . . . . . . . . . . . . . . . 9113.5 Bogoliubov approach . . . . . . . . . . . . . . . . . . . . . . . 94

14 Superconductivity 9714.1 BCS Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . 9714.2 Study of the gap equation . . . . . . . . . . . . . . . . . . . . 10214.3 Finite temperature . . . . . . . . . . . . . . . . . . . . . . . . 10314.4 The BCS ground state . . . . . . . . . . . . . . . . . . . . . . 106

A Fourier transform of the Heaviside distributions 1, δ and θ. 108

B The distribution 1p−i0 109

C Coherent states 110

D Yukawa potential 112

E Dirac equation solutions and their properties 113E.1 Spinors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113E.2 Projection operators . . . . . . . . . . . . . . . . . . . . . . . 116

F Calculation of µ→ eνeνµ decay squared amplitude 118

G Bose Einstein and Fermi Dirac statistics 120G.1 A gas of free fermions . . . . . . . . . . . . . . . . . . . . . . . 124G.2 A gas of free bosons . . . . . . . . . . . . . . . . . . . . . . . . 127

H Fundamental state of the superfluidity theory 128

I Bogoliubov transformation 130

1 Vibrational modes and one dimensional lat-

tice

1.1 Lagrangian and Hamiltonian

Let us consider the elastic vibrations of a cubic cristal. The solution is simplewhen the waves describing the vibrations propagate in the x, y or z directionbecause entire planes of atoms oscillate.

For example for a two dimensional cristal, let us consider a one wavepropagating as k = x. Then one has a longitudinal wave when

uj = x exp [i(kxj − ωt)] (1.1)

where xj = ja being a the lattice size. One has a transverse wave when

uj = y exp [i(kxj − ωt)] (1.2)

In the first case the displacement of the j atom uj is in the x direction whilein the second case is in the y direction. In both cases the problem is reducedto a one dimensional case. Therefore for simplicity we will consider a onedimensional lattice.

Let us consider a one dimensional chain, containing N atoms with spacinga, bound by an elastic force with elastic constant C. Let m be the mass ofthe atoms and xi = ia, i = 1, . . . N , the rest (equilibrium) position of theatoms and x′i the position at the time t. Then the displacement with respectto the equilibrium position is

ui = x′i − xi (1.3)

For the study we will assume boundary periodic conditions

uN+1 = u1 (1.4)

Other conditions could be imposed, for instance fixed boundaries (u1 = 0 =uN+1).

The potential of the chain is given by

V =1

2C(u1 − u2)

2 + (u2 − u3)2 + · · · = 1

2C

N∑i=1

(ui − ui+1)2 (1.5)

1

the kinetic energy

m

2

N∑i=1

u2i (1.6)

Then the Lagrangian is given by

L =m

2

N∑i=1

u2i −1

2C

N∑i=1

(ui − ui+1)2 (1.7)

The lagrangian describes the small oscillations of the atoms with respectto the equilibrium. Higher order terms, cubic or quartic in ui, could beintroduced.

Let us now consider the Euler equations

d

dt

∂L

∂ui=∂L

∂ui(1.8)

or

mui =∂V

∂ui= −C

22∑j

(uj − uj+1)(δi,j − δi,j+1)

= −C(ui − ui+1) + C(ui−1 − ui)

= −C(2ui − ui−1 − ui+1)

= −∑j

Vijuj (1.9)

withVij = C(2δij − δi,j+1 − δi,j−1) (1.10)

This is a finite difference differential equation, which can be solved by adiscrete Fourier transform,

uj(t) = A exp (iχj) exp (−iωχt) + cc (1.11)

By substituting (1.11) in eq.(1.9), we get

m(A exp (iχj) exp (−iωχt) + cc)(−ω2χ) = −C[2A exp (iχj) exp (−iωχt) + cc]

+ C[A exp (iχ(j − 1)) exp (−iωχt) + cc]

+ C[A exp (iχ(j + 1)) exp (−iωχt) + cc]

(1.12)

2

and therefore

ω2χ = 2

C

m(1− cosχ) = 4

C

msin2 χ

2(1.13)

or assuming positive frequency

ωχ = 2

√C

m| sin χ

2| (1.14)

By imposing the condition (1.4) we get

exp (iχ(N + 1)) = exp (iχ) (1.15)

implying

exp (iχN) = 1, χ =2πn

N(1.16)

with −N/2+ 1 ≤ n ≤ N/2 for even N and −(N − 1)/2 ≤ n ≤ (N − 1)/2 forodd N . In conclusion one has N proper modes. Since χ is defined modulus2π, n is defined modulus N . In fact if χ→ χ+ 2mπ, then

n =χN

2π→ χN

2π+

2mπN

2π= n+mN (1.17)

We can rewrite the solution as

uj(t) = A exp (iχj − iωχt) + cc

= A exp (ikxj − iωχt) + cc

= A exp (ikxj − iωnt) + cc (1.18)

with

k =χ

a=

2πn

Na(1.19)

and

ωn = 2

√C

m| sin πn

N| (1.20)

Note that all atoms oscillate with the same frequency ωn. The group velocity

vg =dω

dk=

√C

ma cos

πn

N(1.21)

Furthermore the condition (1.19) gives

nλ = n2π

k= Na (1.22)

or the total length must contain an integer number of wave lengths.

3

1.2 Expansion in eigenmodes and quantization

Since the u(n)i are a set of orthonormal and complete functions, the general

solution can be written in terms of the normal modes, defined as

uj(t) ∼ u(n)j exp (−iωχt) + cc (1.23)

u(n)j =

1√N

exp (i2πn

Nj) (1.24)

Let us show that (1.24) are a set of orthonormal functions, or∑j

(u(n)j )∗u

(m)j = δn,m (1.25)

and satisfy the completeness relation∑n

(u(n)i )(u

(n)j )∗ = δi,j (1.26)

For i = j or n = m the properties are obvious while the orthogonality(completeness) condition follows from the fact that eqs.(1.25 1.26) for n 6= m(i 6= j) become the sum of the Nth roots (or powers of roots) of the identity:∑

j

1

N

[exp(i

2πj

N)

](n−m)

(1.27)

Note that, by inserting eq.(1.23) in the equations of motion (1.9), we get

Viju(n)j = m(ωn)

2u(n)i = m(ωn)

2δiju(n)j (1.28)

which means that the normal modes u(n)i are eigenvectors of the matrix Vij

with eigenvalues m(ωn)2.

The general solution is therefore

uj(t) =∑n

Anu(n)j exp (−iωnt) + c.c. (1.29)

or with a different normalization and notation (ωn → ωk)

uj(t) =∑k

√~

2mωkaku

(k)j exp (−iωkt) + c.c. (1.30)

4

where we recall that ωk = 2√C/m sin(ka/2) with k = 2πn/Na. The mo-

mentum is given by

pj(t) =∂L∂uj

= muj(t) = m∑k

√~

2mωkaku

(k)j (−iωk) exp (−iωkt) + c.c.

(1.31)The Hamiltonian is

H = −L+∑j

ujpj = T + V (1.32)

Let us now invert the relations in order to obtain ak in terms of the coordi-nates and momenta. Let us consider

uj(0) ≡ qj(0) =∑k

√~

2mωk(aku

(k)j + ak

∗(u(k)j )∗) (1.33)

pj(0) =∑k

√mωk~i√2

(aku(k)j − ak

∗(u(k)j )∗) (1.34)

By multiplying by (u(k′)j )∗ and summing over j we get

∑j

(u(k′)j )∗qj(0) =

√~

2mωk′(ak′ + a∗−k′) (1.35)

∑j

(u(k′)j )∗pj(0) =

1

i

√mωk′~

2(ak′ − a∗−k′) (1.36)

from which

ak =∑j

(u(k)j )∗(

√mωk2~

qj(0) + i

√1

2mωk~pj(0)) (1.37)

Let us now quantize the theory by requiring

[qi, pj] = i~δij, [qi, qj] = [pi, pj] = 0 (1.38)

By using eq.(1.37) and eq. (1.38), we get

[ak, a†k′ ] = δkk′ , [ak, ak′ ] = [a†k, a

†k′ ] = 0 (1.39)

5

Since the Hamiltonian does not depend explicitly on t we can quantize att = 0

H(t = 0) =1

2m

∑j

p2j(0) +1

2

∑l,j

Vljql(0)qj(0) (1.40)

where

Vlj =∂V

∂ul∂uj= C(2δl,j − δl,j−1 − δl,j+1) (1.41)

By substituting in the Hamiltonian pj(0) and qj(0) we get

H =1

2

∑k

~ωk(a†kak + aka†k) (1.42)

In fact we have

~2m

∑j

p2j(0) =~2m

∑j

∑k

∑k′

√mωk

i√2

√mωk′

i√2

(aku(k)j − ak

†(u(k)j )∗)(ak′u

(k′)j − ak′

†(u(k′)j )∗)

= −~4

∑k

∑k′

√ωk

√ωk′ [δk,−k′

(akak′ + a†ka†k′)− δk,k′(aka

†k′ + a†kak′)]

= −~4

∑k

ωk(aka−k − aka†k − a†kak + a†ka

†−k) (1.43)

Furthermore we have

1

2

∑l,j

Vljql(0)qj(0) =~2

∑l,j

Vlj∑k

∑k′

1√2mωk

1√2mωk′

(aku(k)l + ak

†(u(k)l )∗)(ak′u

(k′)j + ak′

†(u(k′)j )∗)

=~2

∑k

∑k′

1√2mωk

1√2mωk′

mω2k

[δk,−k′(akak′ + a†ka†k′) + δk,k′(aka

†k′ + a†kak′)]

=~4

∑k

ωk(aka−k + aka†k + a†kak + a†ka

†−k) (1.44)

where use has been made eqs. (1.28) Then the Hamiltonian is given by

H =~2

∑k

ωk(a†kak + aka

†k) =

∑k

~ωk(a†kak +1

2) (1.45)

6

The Hamiltonian is then the sum of N armonic oscillator Hamiltonians. TheHilbert space is built starting from the fundamental state (in quantum fieldtheory this state is called the vacuum) |0 >= |0 >kmin

· · · |0 >kmax such that

ak|0 >= 0, ∀k (1.46)

Remember k = 2πn/Na with (for example for even N)

kmin =2π(−N

2+ 1)

Na, kmax =

2πN2

Na(1.47)

The statea†k|0 > (1.48)

eigenvector of the Hamiltonian with energy ωk represents a possible quantumexcitation of the lattice and it is called phonon. The generic state is givenby

(a†k1)Nk1

Nk1!

(a†k2)Nk2

Nk2!. . . |0 > (1.49)

In conclusion the energy of an atom lattice is quantized. The phonon does notcarry physical momentum however interacts with particles such as photons,neutrons as if it had a momentum k.

1.3 Continuum limit

The continuum limit of the oscillator lattice is obtained by taking the limitN → ∞ or a→ 0 considering L = aN finite.

In this limit xj = ja→ x, k = χ/a is the wave vector.

ui(t) = x′i − xi → u(x, t) (1.50)

ui+1 − uia

→ u′(x, t) (1.51)

where

u′(x, t) =∂u(x, t)

∂x(1.52)

7

The lagrangian (1.7) in the continuum limit becomes

L =m

2

N∑i=1

u2i −1

2C

N−1∑i=1

(ui − ui+1)2

=m

2a

N∑i=1

au2i −1

2Ca2

N−1∑i=1

(ui − ui+1)2

a2

→ µ

2

∫ L

0

u(x, t)2 − 1

2K

∫ L

0

u′(x, t)2 (1.53)

or

L =

∫ L

0

L, L =µ

2u(x, t)2 − 1

2Ku′(x, t)2 (1.54)

where we have introduced the linear density µ = m/a and the comprimibilitymodulus K = Ca. L is called Lagrangian density. Then the equations ofmotion (1.9) become the D’Alembert equation in one dimension

µd2

dt2u = Ku′′ (1.55)

where u′ = du/dx. The velocity is given by

v =

√K

µ(1.56)

In fact in the continuum limit we have

ω2k = 4

K

µa2sin2(

ka

2) → v2k2 (1.57)

The expansion becomes

uj(t) =∑k

√~

2µωkaaku

(k)j exp (−iωkt) + c.c.

=∑k

√~

2µωkLak exp (−i(ωkt− kxj)) + c.c.

→∑k

√~

2µωkLak exp (−i(ωkt− kx)) + c.c. (1.58)

8

where the sum is extended over

k =2πn

Na=

2πn

L(1.59)

with n = 0,±1,±2 · · ·.

The Hamiltonian density is given by

H = Πu− L =1

2µΠ2 +

1

2K(u′)2 (1.60)

where

Π =∂L∂u

(1.61)

The total energy associated to the field is

H =

∫d3xH (1.62)

The quantization of the system is obtained by requiring the commutationrelation at equal times

[u(t, x),Π(t, y)] = i~δ(x− y) (1.63)

[u(t, x), u(t, y)] = [Π(t, x),Π(t, y)] = 0 (1.64)

which imply for the operators ak, a†k

[ak, a†k′ ] = δkk′ (1.65)

[ak, ak′ ] = [a†k, a†k′ ] = 0 (1.66)

The total Hamiltonian, using the expansion (1.58), becomes

H =∑k

~ωk(a†kak +1

2) (1.67)

Since the operator a†kak is definite positive, a state with minimum energy|0 > exists, defined by

ak|0 >= 0, ∀k (1.68)

9

This state corresponds to the fundamental state or of minimum energy. Thegeneric state of the Hilbert space is built as a linear combination of the states(Fock states)

(a†k1)nk1

√nk1!

(a†k2)nk2

√nk2!

. . . (|0 >) (1.69)

The total energy of this state is the sum of the energy ωki of the singlequantum excitations

E =∑i

nkiωki (1.70)

2 Lagrangian field theory

2.1 Action and Euler equations for a continuous sys-tem

In general the action for a continuous system is given by:

S =

∫ tf

ti

dt

∫V

d3xL (2.1)

where L is the Lagrangian density with

L = L(φA(t,x), φA(t,x), ∂iφA(t,x)), i, j = 1, 2, 3 (2.2)

φA(t,x), A = 1 . . . q are q fields and ∂if ≡ ∂f/∂xi. For lagrangians depend-ing on higher derivatives see [1] (Ostrogadski method).

The Action Principle requires the stationarity of the action for any vari-ation δφA such that

δφA|∂V = 0, δφA|ti = δφA|tf = 0 (2.3)

Therefore (from now on the sum over equal indices is implied) we obtain

0 = δS =

∫dtd3x(

∂L∂φA

δφA +∂L∂φA

δφAi +∂L

∂∂jφAδ∂jφA)

=

∫dtd3x(

∂L∂φA

− ∂

∂t

∂L∂φA

− ∂j∂L

∂∂jφA)δφA

10

+

∫dtd3x(

∂

∂t(∂L∂φA

δφA) + ∂j(∂L

∂∂jφAδφA))

=

∫dtd3x(

∂L∂φA

− ∂

∂t

∂L∂φA

− ∂j∂L

∂∂jφA)δφA

(2.4)

which implies the Euler equations

∂

∂t

∂L∂φA

+ ∂j∂L

∂∂jφA=

∂L∂φA

(2.5)

It is easy using (2.5) to get eqs.(1.55) from the lagrangian given in eq. (1.54)

In the relativistic case the lagrangian

L = L(φA(x), ∂µφA(x)) (2.6)

where µ = 0, 1, 2, 3 and the Euler equations (2.5) become

∂

∂xµ∂L∂µφA

=∂L∂φA

(2.7)

The sum over µ is understood.

Example. Schrodinger field. The Lagrangian for the Schrodingerfield is given by (assuming ~ = 1):

L =i

2(ψψ∗ − ψψ∗)− 1

2m∇ψ∗∇ψ − V ψ∗ψ (2.8)

3 Quantization of the Klein-Gordon field

3.1 Quantization of the Klein-Gordon field in 1D

Let us consider the Klein-Gordon1 real field in 1D, x ∈ [0, L], obtained fromthe continuous string lagrangian, eq.(1.54), assuming µ = 1, K = 1 and

1The equation in 3D was proposed indipendently by O. Klein, W. Gordon, V. Fockand E. Schrodinger in 1926

11

adding a quadratic mass term. We assume the light velocity c = 1 and~ = 1.

L =1

2

(φ2 − φ′2 −m2φ2

)(3.9)

From the Euler equations we get

(∂2

∂t2− ∂2

∂x2+m2)φ = 0 (3.10)

The solutions are, using the expansion (1.58) with µ = 1, ωk ≡ Ek

φ(t, x) =∑k

√1

2EkL[ak exp (−i(Ekt− kx)) + h.c.] (3.11)

with k = 2πn/L, n = 0,±1,±2 . . . and

Ek =√k2 +m2 (3.12)

The momentum density is given by

Π =∂L∂φ

= φ (3.13)

The Hamiltonian is defined

H =

∫dxH (3.14)

with

H = Πφ− L =1

2[φ2 + φ′2 +m2φ2] (3.15)

In order to quantize the theory we promote φ and Π to self adjoint oper-ators satisfying “equal time” commutation relations

[φ(t, x),Π(t, y)] = iδ(x− y) (3.16)

[φ(t, x), φ(t, y)] = [Π(t, x),Π(t, y)] = 0 (3.17)

We can now compute the commutation relations between the operators akand a†k. Before let us show that

ak =1√2LEk

∫ L

0

dx exp (i(Ekt− kx))[Ekφ+ iφ] (3.18)

12

In fact we have:

1√2LEk

∫ L

0

dx exp (i(Ekt− kx))[Ek

∑k′

√1

2Ek′L[ak′ exp (−i(Ek′t− k′x))

+a†k′ exp (i(Ek′t− k′x))]

+∑k′

√1

2Ek′LEk′ [ak′ exp (−i(Ek′t− k′x))

−a†k′ exp (i(Ek′t− k′x))]]

=1

2L√Ek

∫ L

0

dx∑k′

1√Ek′

[(Ek + Ek′)ak′ exp (i((Ek − Ek′)t− (k − k′)x))

+(Ek − Ek′)a†k′ exp (i((Ek + Ek′)t− (k + k′)x))

=1

2

∑k′

1√Ek′Ek

[(Ek + Ek′)ak′δk,k′ exp (i(Ek − Ek′)t)

+(Ek − Ek′)a†k′δk,−k′ exp (i(Ek + Ek′)t)]

= ak (3.19)

where use has been made of

1

L

∫ L

0

dx exp [i(k − q)x] = δkq (3.20)

Using (3.18) and (3.16)-(3.17), we have

[ak, a†k′ ] = δkk′

[ak, ak′ ] = [a†k, a†k′ ] = 0 (3.21)

Using the expansion for φ, given in eq.(3.11), and the commutation relations,we get

H =∑k

Ek(a†kak +

1

2) (3.22)

The generic state of the Hilbert space is built as a linear combination ofthe states (Fock states)

(a†k1)nk1

√nk1!

(a†k2)nk2

√nk2!

. . . (|0 >) (3.23)

13

and contains nk1 particles with energy Ek1, nk2 particles with energy Ek2...The fundamental state |0 > is such that

ak|0 >= 0, ∀k (3.24)

The state |0 > is called also the vacuum state because no particles are present.Note that the vacuum expectation value of the Hamiltonian is infinite:

< 0|H|0 >= 1

2

∑k

Ek (3.25)

However this is not a problem since usually one measures differences of en-ergies between states.

3.2 Quantization of Klein-Gordon field in 3D

The action for the Klein-Gordon field in three spatial dimensions is obtainedby generalizing the previous section

S =

∫ tf

ti

dt

∫V

d3x1

2

(φ2 − (∇φ)2 −m2φ2

)(3.26)

where V ≡ L3. This action can be written also in the covariant form

S =

∫V4

d4x1

2

(∂µφ∂

µφ−m2φ2)

(3.27)

where V4 is now a space-time volume, d4x is the space time volume elementand µ = 0, 1, 2, 3. The sum over the repeated indices µ is understood. Assum-ing the mass as fundamental dimension, the dimension of the Klein-Gordonfield ar M1 so that the action is dimensionless. From the Euler equations weget the differential equations

(∂2

∂t2−∇2 +m2)φ = 0 (3.28)

or in covariant form

(∂µ∂µ +m2)φ ≡ (+m2)φ = 0 (3.29)

14

This equation is the simplest relativistic estension of the Schrodinger equa-tion, obtained by replacing the relativistic mass condition

(E2k − k2 −m2) = 0 (3.30)

with the differential equation 3.28 obtained by substituting the classical vari-ables Ek and k by the operators

Ek → i∂

∂t, k → −i∇ (3.31)

As we will see the quantization of this field theory describe a many bosonrelativistic theory.

To quantize this theory, we postulate equal time commutation relationsbetween the operators φ and Π are

[φ(t,x),Π(t,y)] = iδ3(x− y) (3.32)

[φ(t,x), φ(t,y)] = [Π(t,x),Π(t,y)] = 0 (3.33)

The field expansion is now

φ(t,x) =∑k

√1

2EkL3[ak exp (−i(Ekt− k · x)) + h.c.] (3.34)

with the dispersion relation of a relativistic particle

Ek =√k2 +m2 (3.35)

and

ki =2πniL

, i = 1, 2, 3, ni = 0,±1,±2, . . . (3.36)

The inversion relation for the operators ak is now

ak =1√

2L3Ek

∫V

d3x exp (i(Ekt− k · x))[Ekφ+ iφ] (3.37)

and the commutations relations are trivial generalization of eqs.(3.21).

The Hamiltonian is

H =

∫d3xH =

∫d3x

1

2[φ2 + (∇φ)2 +m2φ2] (3.38)

15

and using the field expansion it turns out to be

H =∑k

Ek(a†kak +

1

2) (3.39)

The Hilbert space is built as in the previous section, with suitable gener-alizations. For instance the state with Nk particles with energy Ek is givenby

(a†k)nk

√nk!

|0 > (3.40)

As we will see in the following these particles have momentum k and spin 0(bosons).

Let us finally note that when working in R3, the expansion becomes aFourier transform

φ(x) =1

(2π)3/2

∫d3k

1√2Ek

[a(k)e−ikx + h.c.

](3.41)

where now x and k denote, using the covariant notations, the fourvectorsand

kx = k0x0 − k · x ≡ Ekt− k · x (3.42)

3.3 Noether Theorem

Let us now prove the Noether theorem2. The theorem states that to everycontinuous transformation for which the action is invariant (δS = 0), therecorresponds a definite invariant which is time conserved. The invariant is thegenerator of the corresponding infinitesimal transformation. Let us considera generic action (we use natural units, ~ = c = 1)

SV =

∫V

d4xL(φA, ∂µφA) A = 1, · · ·N (3.43)

where V is a space-time volume.We are going to consider both a variation ofform of the fields and of the space-time coordinates

φA → φA, x→ x′ (3.44)

2Emmy Noether, 1882-1935

16

In general he total variation of a function is defined as

∆f = f ′(x′)− f(x) = f ′(x′)− f(x′) + f(x′)− f(x) (3.45)

For infinitesimal transformations we have

∆f = δf +∂f

∂xδx (3.46)

where we have defined the local variation

δf = f ′(x)− f(x) (3.47)

Let us suppose that the action is invariant under (3.44)

S ′V ′ =

∫dx′L′(x′) = SV (3.48)

whereL′(x′) = L′(φA(x

′), ∂′µφA(x′)) = L(x) + ∆L(x) (3.49)

with

∆L(x) ∼ ∂L∂φA

δφA +∂L

∂∂µφAδ∂µφA +

∂L∂xµ

δxµ (3.50)

with δφA and δ∂µφA the local variations

δφA = φ′A(x)− φA(x) (3.51)

The variation of the action can be written as

δSV =

∫V ′d4x′L′(x′)−

∫V

d4xL(x) (3.52)

or

δSV =

∫V ′d4x′L(x) +

∫V ′d4x′∆L(x)−

∫V

d4xL(x)

∼∫V

d4xL(x)|∂x′

∂x|+

∫V

d4x∆L(x)−∫V

d4xL(x)

∼∫V

d4xL(x)(1 + ∂µδxµ)−

∫V

d4xL(x)

+

∫V

d4x[∂L∂φA

δφA +∂L

∂∂µφAδ∂µφA +

∂L∂xµ

δxµ]

∼∫V

d4x[∂L∂φA

− ∂µ(∂L

∂∂µφA)]δφA

+

∫V

d4x∂µ[∂L

∂∂µφAδφA + Lδxµ] (3.53)

17

In the previous equations the Jacobian has been computed as

det(I +∆) = expTr ln(1 + ∆) ∼ expTr∆ ∼ 1 + Tr∆ (3.54)

with∆µν = ∂νδx

µ (3.55)

Using the Euler equations of motion in (3.53), we obtain∫V

d4x∂µ[∂L

∂∂µφAδφA + Lδxµ] = 0 (3.56)

By considering the global variation

∆φA = φA(x′)− φA(x) (3.57)

for infinitesimal transformation we have

∆φA ∼ δφA + ∂µφAδxµ (3.58)

In term of the global variation ∆φA we get∫V

d4x∂µ[∂L

∂∂µφA∆φA + Lδxµ − δxν

∂L∂∂µφA

∂νφA] = 0 (3.59)

Let us now consider the consequences of the Noether theorem in someexamples.

Space-time Translations Let us consider an infinitesimal space timetranslation

δxµ = εµ (3.60)

Under this transformation the fields are invariant

∆φA = 0 (3.61)

From eq.(3.59) we get

∂µ[Lεµ − εν∂L

∂∂µφA∂νφA] = 0 (3.62)

18

Being εµ arbitrary we get∂µT

µν = 0 (3.63)

where

T µν =∂L

∂∂µφA∂νφA − gµνL (3.64)

By integrating over d3x the µ = 0 component of T µν , we get four invariantscorresponding to the total four-momentum of the field

P ν =

∫d3xT 0ν (3.65)

In particular for ν = 0 we recover the Hamiltonian. For ν = i, we get thespatial momentum associated to the field

P i =

∫d3x

∂L∂φA

∂iφA (3.66)

For example, in the case of the Klein-Gordon field,

∂L∂φ

= φ (3.67)

and

P i =

∫d3xφ∂iφ =

∑k

kia†kak (3.68)

In conclusion each quantum of the Klein-Gordon field contributes to themomentum of the field with its momentum k.

Lorentz Transformations Let us then consider an infinitesimal Lorentztransformation

Λµ·ν = δµν + εµ·ν (3.69)

The matrix conditionΛTgΛ = g (3.70)

implies that the tensor εµν is antisymmetric,

εµν = −ενµ (3.71)

19

Let as assume that under the Lorentz transformation the fields transform as

∆φA = −1

2ΣµνABεµνφB (3.72)

Let us consider some examples. For the Klein-Gordon field we have

φ′(x′) = φ(x(x′)) (3.73)

or the field is a scalar field. For vector fields, like the four potential Aµ(x),wehave

A′µ(x′) = ΛµνAν(x(x′)) (3.74)

By substituting (3.72) in (3.59), we obtain

0 = ∂µ[−∂L

∂∂µφA

1

2ΣρσABερσφB + Lεµσxσ − ενσxσ

∂L∂∂µφA

∂νφA]

= ∂µ[−∂L

∂∂µφA

1

2ΣρσABερσφB + Lερσgµρxσ − ερσx

σ ∂L∂∂µφA

∂ρφA]

= ∂µ[−∂L

∂∂µφA

1

2ΣρσABερσφB − ερσx

σT ρµ]

= ∂µ[−∂L

∂∂µφA

1

2ΣρσABερσφB − 1

2ερσ(x

σT ρµ − xρT σµ)]

(3.75)

or∂µMµρσ = 0 (3.76)

with

Mµρσ = xρT σµ − xσT ρµ − ∂L∂∂µφA

ΣρσABφB (3.77)

In conclusion we can built the six invariants

Mρσ =

∫d3xM0ρσ (3.78)

M ij are the angular momentum components or the generators of the O(3)rotations, while M0i are the generators of the Lorentz transformations.

20

Internal transformations In this case only the field transforms, δx = 0,∆φ = δφ. Therefore from eq. (3.59), we get

∂µ∂L

∂∂µφA∆φA = 0 (3.79)

As an example let us consider the gauge transformations for the SchrodingerLagrangian

L =i

2(φ∗φ− φ∗φ)− 1

2m∇φ∗∇φ− V φ∗φ (3.80)

The lagrangian is invariant under the internal transformation

φ→ exp(iα)φ (3.81)

where α is a real number. Therefore for infinitesimal transformation

∆φ ∼ iαφ (3.82)

The Noether theorem implies∫d4x

[(∂

∂t

∂L∂φ

+ ∂k∂L∂∂kφ

)∆φ+ (φ→ φ∗)

]= 0 (3.83)

But∂L∂φ

=i

2φ∗,

∂L∂φ∗

= − i

2φ (3.84)

∂L∂∂kφ

= − 1

2m∂kφ

∗,∂L∂∂kφ∗ = − 1

2m∂kφ (3.85)

and substituting in (3.83), we get

α

∫d4x

[∂

∂t(−φ∗φ) + ∂k(−

i

2m∂kφ

∗φ+i

2mφ∂kφ

∗)

]= 0 (3.86)

or [∂

∂t(−φ∗φ) + ∂k(−

i

2m∂kφ

∗φ+i

2mφ∂kφ

∗)

]= 0 (3.87)

In conclusion, we obtain the continuity equation[∂

∂t(φ∗φ) +∇(+

i

2m(∇φ∗)φ− i

2mφ∇φ∗)

]= 0 (3.88)

21

4 Quantization of the electromagnetic field

4.1 Lagrangian and Hamiltonian

Let us start by writing the lagrangian of the electromagnetic field

L = −1

4FµνF

µν =1

2F0iF

0i − 1

4FijF

ij =1

2(E2 −B2) (4.1)

where we have used F 0i = −Ei and F ij = −εijkBk. From this Lagrangianwe can derive the Euler equations

∂L∂Aρ

= ∂σ

(∂L

∂∂σAρ

)(4.2)

Now∂L

∂∂σAρ=

∂L∂∂Fµν

∂Fµν∂∂σAρ

= −F σρ (4.3)

and therefore we get the free equations of motion

∂σFσρ = 0 (4.4)

In presence of an interaction with a corrent µ

Ltot = L+ LI (4.5)

LI = −jµAµ (4.6)

and the Euler equations of motion are the Maxwell equations in presence ofthe current jµ

∂σFσρ = jρ (4.7)

To quantize the system, let us first compute the Hamiltonian. Let define themomentum density as

Πµ =∂L∂Aµ

(4.8)

Using the eq. (4.3), we getΠµ = −F0µ (4.9)

Therefore the zero component Π0 = 0 and

Πi = −F0i = −Ei = Ai + ∂iA0 (4.10)

22

The Hamiltonian densityH = ΠiA

i − L (4.11)

and the Hamiltonian

H =

∫d3xH =

1

2

∫d3x(Π2 +B2)−

∫d3xΠi∂iA

0 (4.12)

The last term can be integrated by parts, neglecting the boundary terms,

−∫d3xΠi∂iA

0 =

∫d3x∂iΠiA

0 = 0 (4.13)

where use has been done of∇ · E = 0 (4.14)

The quantization of the electromagnetic field can be performed in differentgauges. Here we choose the Coulomb gauge

∇ ·A = 0 (4.15)

In this gauge∇ · E = 0 = ∂iA

i +4A0 = 4A0 (4.16)

In conclusion, eq.(4.1) implies A0 = 0. The gauge

∇ ·A = 0, A0 = 0 (4.17)

is called radiation gauge. Working in this gauge we loose manifest covariance.The advantage is that we work with the two independent degrees of freedom,as we will see when expanding the field in normal modes. A0 and Π0 will notbe quantized and the field A is transverse, due to eq.(4.17).

We can now expand the field in normal modes in a finite volume V

A(x) =∑k

∑α=1,2

1√2V ωk

εαk[aαke

−ikx + h.c.], ki = ni2π

L(4.18)

wherekx ≡ k0x0 − k · x (4.19)

and k0 ≡ ωk. The polarization vectors εαk are orthogonal and transverse tok. For simplicity we assume real polarization vectors.

εαk · εβk = δαβ, εαk · k = 0 (4.20)

23

The Hamiltonian is

H =1

2

∫d3x[Π2 + (∇×A)2] (4.21)

whereΠ = A (4.22)

The quantization3 is obtained by requiring the commutation relations

[aαk, aβ†k′ ] = δαβδk,k′ (4.23)

[aαk, aβk′ ] = [aα†k , a

β†k′ ] = 0 (4.24)

By substituting in the Hamiltonian the expansion in normal modes (4.18)we get

H =∑k

∑α=1,2

ωk(aα†k a

αk +

1

2) (4.25)

Using the Noether theorem we can build the energy momentum tensor

T 0i =

∂L∂Aµ

∂iAµ =

∂L∂Ak

∂iAk = Πk∂iA

k = Ak∂iAk (4.26)

Therefore the momentum of the field is

P i =

∫d3xT 0i = −

∫d3xAk∂iA

k (4.27)

By substituting the expression in normal modes (4.18), we obtain

P =∑k

∑α=1,2

k(aα†k aαk) (4.28)

3The first attempt of quantization of the electromagnetic field was performed by M.Born, W. Heisenberg and P. Jordan in 1926. Then in 1927 Dirac published the paper onThe quantum theory of the emission and absorption of radiation. The idea of the secondquantization was also proposed by Jordan, in 1927, while the expression was coined byDirac. The general theory of quantum fields through the method of canonical quantizationwas presented in W. Heisenberg and W. Pauli in 1929.

24

Note that the usual expression of the Poynting vector is equivalent to(4.26), since∫

d3x(E×B)i =

∫d3xεijkA

jεklm∂lAm

=

∫d3xAj∂lA

m[δilδjm − (l ↔ m)]

=

∫d3x(Am∂iA

m − Al∂lAi)

=

∫d3xAm∂iA

m (4.29)

where in the last step we have eliminated a term by integrating by parts andtaking into account the Coulomb gauge condition.

The Hilbert space is built by many photons states

|nk1,α1 , nk2,α2 . . . > (4.30)

with

|nk1,α1 , nk2,α2 . . . >=(a†k1,α1)

)nk1,α1√nk1,α1 !

(a†k2,α2))nk1,α1√

nk2,α2 !. . . |0 > (4.31)

withak,α|0 >= 0, ∀α,k (4.32)

In conclusion the total energy and the total momentum of the electromagneticfield ar the sum of single photon contributions of energy ωk and momentumk.

H|nk1,α1 , nk2,α2 . . . >=∑i

ωkinki,αi

|nk1,α1 , nk2,α2 . . . > (4.33)

P|nk1,α1 , nk2,α2 . . . >=∑i

kinki,αi|nk1,α1 , nk2,α2 . . . > (4.34)

Therefore the mass of the photon is zero

m2 = ω2k − k2 = 0 (4.35)

The photon is characterized by the momentum k and by the polarizationvector. From the two one photon states

a†k,1|0 >, a†k,2|0 > (4.36)

25

we can define the circular polarization states

∓ 1√2[a†k,1 ∓ a†k,2]|0 > (4.37)

These two states are eigenvectors of the helicity operator, the projection ofthe spin of the photon along the photon flight direction), with eigenvalues±1 (see for example [2, 17]) . In conclusion the photon has spin one but onlystates with helicity ±1 are allowed. This is related to the fact that the photonis massless and it is a consequence of the theory of Poincare representations.

Using the expansion (4.18) and the commutation relations (4.24) we cancompute the commutation relations between the canonical operators fieldand momentum density:

[Ai(t,x),Πj(t,y)] =i

V

∑k,α

(δij − kikj

k2)eik·y

= i[δijδ3(x− y)− ∂i∂j

∇2δ3(x− y)]

= i[δijδ3(x− y) + ∂i∂j1

4π|x− y|] (4.38)

[Ai(t,x), Aj(t,y)] = [Πi(t,x),Πj(t,y)] = 0 (4.39)

The three A operators are not independent. since we are quantizing inthe Coulomb gauge. As a consequence the commutator between the field andthe momentum density is not canonical (see [3]).

4.2 Casimir effect

The Casimir effect is the macroscopic manifestation of the vacuum fluctua-tions of the quantized electromagnetic field:

< 0|∫d3x[E2 +B2]|0 >= 1

2

∑α,k

ωk (4.40)

H. Casimir (1909-2000), a dutch physicist, wrote the paper on this effectin 1948. This effect was experimentally detected in 1958 by Spaarnay andchecked with percent accuracy by Lamoreaux (1998).

26

We will show that the vacuum energy (4.40) generates an actractive forcebetween the faces of metallic plates at a distance d, such that the force forsurface unit

f(d) =F (d)

L2∼ hc

d4(4.41)

Let us consider two metallic squared parallel plates at distance d insidea the cubic box of volume L3. Let us assume that the conducting plates areorthogonal to the z axis. The electric field in the vacuum is solution of theMaxwell equation with the boundary condition that the tangential field Etg

has to vanish on the conducting wall (z = 0, d).

Therefore the tangential field behaves as

Etg ∼ sin(kzz) (4.42)

withkz =

nπ

d, with n = 1, 2... (4.43)

The energy is

ωk =

√k2x + k2y + (

nπ

d)2 (4.44)

withkx,y =

nx,yπ

L, nx,y = −∞, ...∞ (4.45)

So the total vacuum energy is

U(d) = 21

2

∑kx,ky ,n

ωk (4.46)

where the 2 comes from the polarization sum. Let us define k =√k2x + k2y

so that kdk = ωkdωk ≡ ωdω where

ω =

√k2 +

(nπd

)2

(4.47)

and pass to the continuum (the plaque lenght L→ ∞):

U(d) = L2

∞∑n=1

∫d2k

(2π)2ω

27

=L2

(2π)2

∞∑n=1

∫ ∞

0

kdk

∫ 2π

0

dϕω

=L2

2π

∞∑n=1

∫ ∞

nπ/d

dωω2 (4.48)

The integral is divergent; there are several ways to regularize the energy toget finite physical results. The result does not depend on the chosen regular-ization (see for example [4]). Let us consider, by introducing a convergencefactor exp(−εω),

U(d, ε) =L2

2π

∞∑n=1

∫ ∞

nπ/d

dωω2e−εω

=L2

2π

d2

dε2

∞∑n=1

∫ ∞

nπ/d

dωe−εω

=L2

2π

d2

dε2

∞∑n=1

e−εnπ/d

ε

=L2

2π

d2

dε2

[1ε(

1

1− e−πε/d− 1)

](4.49)

The energy for surface unit is given by

U(d, ε)

L2=

1

2π

d2

dε2

[1ε(

1

1− e−πε/d− 1)

](4.50)

To evaluate the limit for ε→ 0 we need the following series

1

1− et= −

∞∑n=0

Bntn−1

n!(4.51)

where Bn are the Bernouilli numbers (B0 = 1, B1 = −1/2, B2 = 1/6, B3 =0, B4 = −1/30). The Bernouilli numbers can be evaluated by expanding inTaylor series

z

1− ez=

∞∑n=0

Bnzn

n!(4.52)

Therefore we obtain

U(d, ε)

L2= − 1

2π

d2

dε2

[1ε[1 +

∞∑n=0

Bn(−πε/d)n−1

n!

]28

= − 1

2π

d2

dε2

[1ε[1−B0

d

πε+B1 −B2

πε

2d−B4

π3ε3

24d3+O(ε4)]

](4.53)

Neglecting terms that do not contribute for ε→ 0 we get

U(d, ε)

L2= 3

B0d

π2ε4− (1 +B1)

1

πε3+B4π

2

24d3

= C0d+ C1 +C2

d3(4.54)

In conclusion we have the C2 finite term which gives the correct result

f(d) = − ∂

∂d

U(d, ε)

L2= − π2

240d4(4.55)

but still two divergent terms, C0, C1 when ε → 0. The second C1 does notcontribute to the force, so we can put C1 = 0. The second can be treatedwith a trick, which consists in introducing two additional external plates atdistance 2D so that we end with three condensators. The total energy forsurface unit is now

U(d,D, ε)

L2= 2[C0(D − d/2) +

C2

(D − d/2)3] + C0d+

C2

d3

= 2C0D + C2(1

d3+

2

(D − d/2)3) (4.56)

Then

f(d,D) = − ∂

∂d

U(d,D, ε)

L2= −3C2

d4+ C2(−3)

1

(D − d/2)4(−1/2) (4.57)

and taking the limit D → ∞

limD→∞

f(d,D) = − π2

240d4(4.58)

We have obtained the result in the system of natural units. The di-mensions of the force for unit surface f are L−4. In the cgs units we have[f ] =ML−1T−2. On the other hand

[~c] =ML3T−2 (4.59)

29

Therefore to get the right dimensions in the cgs system, we have to multiplyby ~c

F

A= − π2

240

~cd4

(4.60)

For d = 10µm, we obtain a tiny force

F

A= −1.3× 10−6dyne/cm2 (4.61)

The gravitational force for unit area of two plates of m = 1g and L =1cm at a distance of 10µm is comparable. However at smaller distances theCasimir force becomes dominant.

5 Hamiltonian for a system of non relativistic

charged particles interacting through the

electromagnetic field

In this chapter we consider several processes of interaction between matterand the electromagnetic fields, like emission and absortion of photons byatoms, the scattering of photons over atoms and the emission of light bycharged particles (Cherenkov effect4). In order to compute the processes ofemission and absortion of photons by atoms we need to consider a systemof non relativistic charged particles interacting through the electromagneticfield.

The Lagrangian for a system of N non relativistic particles with mass mr

and charge er in interaction between themselves through the electromagneticfield is given by

L =∑r

[12mrξ

2r − erA0(ξr, t) + erξr ·A(ξr, t)

]+

∫d3xL (5.1)

with

L =1

2(E2 −B2) (5.2)

4P. Cherenkov (1904-1990), Nobel prize in Physics in 1958

30

Let us perform the Lagrange transform. Defining

pr =∂L

∂ξr= mrξr + erA(ξr, t)

Πi =∂L∂A

i = −Ei = ∂iA0 + Ai (5.3)

the Hamiltonian is

H =∑r

pr(t) · ξr(t) +∫d3xΠ · A(x, t)− L (5.4)

We obtain (from now on ξr ≡ ξr(t),pr ≡ pr(t))

H =∑r

1

2mr

(pr − erA(ξr, t))2 +

∑r

erA0(ξr, t)

+

∫d3x

1

2(E2 +B2)−

∫d3xΠi∂iA0(x, t)

=∑r

1

2mr

(pr − erA(ξr, t))2 +

∑r

erA0(ξr, t)

+

∫d3x

1

2(E2 +B2) +

∫d3x∂iΠiA0(x, t) (5.5)

where we have integrated by parts. On the other hand

∂iΠi = −∂iEi = −ρ(x, t) = −∑r

erδ(x− ξr) (5.6)

So

H =∑r

1

2mr

(pr − erA(ξr, t))2 +

∑r

erA0(ξr, t)

+

∫d3x

1

2(E2 +B2)−

∑r

er

∫d3xδ(x− ξr)A0(x, t)

=∑r

1

2mr

(pr − erA(ξr, t))2 +

∫d3x

1

2(E2 +B2)

=∑r

1

2mr

(pr − erA(ξr, t))2

31

+

∫d3x

1

2[(A(x, t)2 + (∇A0(x, t))

2 + 2A(x, t) · ∇A0(x, t) + (∇×A(x, t))2]

=∑r

1

2mr

(pr − erA(ξr, t))2

+

∫d3x

1

2[(A(x, t)2 − (∇2A0(x, t)A0(x, t)) + (∇×A(x, t))2]

=∑r

1

2mr

(pr − erA(ξr, t))2

+

∫d3x

1

2[(A(x, t)2 +

∑r

erδ(x− ξr)A0(x, t) + (∇×A(x, t))2]

=∑r

1

2mr

(pr − erA(ξr, t))2 +

1

2

∑r 6=s

eres4π|ξr − ξs|

+

∫d3x

1

2[(A(x, t)2 + (∇×A(x, t))2]

(5.7)

where we have usedA0(x, t) =

∑s 6=r

es4π|ξr − ξs|

(5.8)

In conclusion the Hamiltonian is given by

H = Hatom +Hrad + V (5.9)

where

Hatom =∑r

p2r

2mr

+ Vcoul (5.10)

is the atomic Hamiltonian with

Vcoul =1

2

∑r 6=s

eres4π|ξr − ξs|

(5.11)

and

V = −∑r

ermr

pr ·A(ξr, t) +∑r

er2

2mr

(A(ξr, t))2

≡ V1 + V2 (5.12)

32

V is the interaction between the charged particles and the electromagneticfield. Hrad denotes the Hamiltonian of the electromagnetic field.

Then we perform the first quantization of the atomic system and thequantization of the electromagnetic field:

[ξir, pjs] = iδijδrs (5.13)

[Ai(x, 0), Aj(y, 0)] = i(δijδ3(x− y) + ∂i∂j1

4π|x− y|) (5.14)

or remembering the expansion in normal modes

A(x, 0) =∑α=1,2

∑k

1√2ωkV

(ε(α)k aαke

ik·x + h.c.) (5.15)

using the creation and annihilation operators

[aαk, aβ†k′ ] = δαβδk,k′ (5.16)

We are now able to compute emission and absortion from an atom byconsidering the perturbation theory that we recall in the following sectionand the interaction Hamiltonian given in eq.(5.12). Notice that the generalform has a first term linear in aαk and aα†k , which can describe processes ofemission and absorption of one photon. The second which is bilinear anddescribes processes where the number of photons can change of two or zerophotons.

Other processes could be studied like Thomson, Rayleigh, Raman scat-tering, photoelectric effect, bremstrahlung,..

6 Scattering theory

6.1 S matrix

As we have seen as we progress from the discussion of the previous sectionof the free fields and particles to the more realistic case of field and particlesin interaction, it is much more difficult to find exact solutions to the prob-lem. In these cases, like radiative transitions in atoms, processes of quantum

33

electrodynamics, the solutions can be found only perturbatively, that is byexpanding in power of the interaction strength. In electrodynamics the ex-pansion parameter is the fine structure constant α = e2/4π ∼ 1/137, which issufficiently small to make succesfull the perturbation series. In the theory ofstrong interactions, Quantum ChromoDynamics (QCD), the correspondingparameter αs ∼ 0.1 and therefore the expansion is more problematic.

To develop the perturbative methods it is first convenient to introduce theinteraction (or Dirac) representation. Let us start by recalling the Schrodingerequation and representation

id

dt|φ(t) >S= H|φ(t) >S (6.1)

If H does not depend explicitly on the time, we can build the evolutionoperator

U = e−iH(t−t0) (6.2)

such that|φ(t) >S= U |φ(t0) >S (6.3)

The operator U is unitary, U †U = I. For simplicity from now on we assumet0 = 0.

In the Heisenberg representation operators and states coincide, at t = 0,with the corresponding operators and states of the Schrodinger representa-tion. For general t, if OS denotes the operator in the Schrodinger represen-tation, tthe corresponding operator in the Heisenberg representation is

OH = U †OSU (6.4)

and satisfies

id

dtOH = [OH , H] (6.5)

Let us now assume that the Hamiltonian is given by

H = H0 +HI (6.6)

where H0 is the Hamiltonian in absence of the interaction and HI is theinteraction Hamiltonian. For example in the radiation matter interaction

HI = −∑r

ermr

pr ·A(ξr, t) +∑r

er2

2mr

(A(ξr, t))2 (6.7)

34

Let us define the vectors in the interaction representation as

|φ(t) >I= U †0 |φ(t) >S (6.8)

withU0 = e−iH0t (6.9)

and the operators asOI = U †

0OSU0 (6.10)

The operators satisfy the equation

id

dtOI = [OI , H0] (6.11)

while the states

id

dt|φ(t) >I= HI

I (t)|φ(t) >I (6.12)

withHII (t) = U †

0HIU0 (6.13)

the Hamiltonian in the interaction representation. In fact

id

dt|φ(t) >I = i

d

dtU †0 |φ(t) >S= −H0U

†0 |φ(t) >S +U †

0(H0 +HI)|φ(t) >S

= U †0HI |φ(t) >S= HI

I |φ(t) >I (6.14)

Therefore when the interaction is switched off, the state vector remain con-stant in time. Let us now study the evolution operator in the interactionrepresentation, defining

|φ(t) >I= UI(t, t0)|φ(t0) >I (6.15)

with UI(t0, t0) = I. Using the equation (6.14), we obtain

id

dtU(t, t0) = HI

IUI(t, t0) (6.16)

One of the advantages of the interaction representation is that, whenthe interaction is turned off, the vectors are constant in time. Usually theinteraction is localized in time and one assumes that in the far past and inthe far future the states are eigenstates of H0.

35

So let us assume that, at the time t = ti = −∞, the state is described bythe vector

|φ(−∞) >≡ |i >≡ limt→−∞

|φ(t) >I (6.17)

which is an eigenstate of the H0 Hamiltonian. The S matrix is defined as

|φ(+∞) >= limt→∞

|φ(t) >I= limt→∞,t0→−∞

UI(t, t0)|φ(t0) >≡ S|φ(−∞) >

(6.18)or

S = limt→∞,t0→−∞

UI(t, t0) (6.19)

For any final state |f > one considers the matrix element

< f |φ(+∞) >= Sfi (6.20)

The solution of eq.(6.16) can be obtained in iterative way

UI(t) = I − i

∫ t

t0

HII (t1)dt1 + (−i)2

∫ t

t0

HII (t1)dt1

∫ t1

t0

HII (t2)dt2 . . . (6.21)

The series of the S matrix is given by the so-called Dyson series

S =∞∑n=0

(−i)n∫ ∞

−∞dt1

∫ t1

−∞dt2 . . .

∫ tn−1

−∞dtnH

II (t1)H

II (t2) . . . H

II (tn) (6.22)

In general [HII (ti), H

II (tj)] 6= 0. Note also that the integrals are time ordered,

t > t1 > t2 · · ·.

We can rewrite (6.22) as

S =∞∑n=0

(−i)n 1

n!

∫ ∞

−∞dt1

∫ ∞

−∞dt2 . . .

∫ ∞

−∞dtnT (H

II (t1)H

II (t2) . . . H

II (tn))

(6.23)with

T (HII (t1)) = HI

I (t1) (6.24)

T (HII (t1)H

II (t2)) = θ(t1 − t2)H

II (t1)H

II (t2) + θ(t2 − t1)H

II (t2)H

II (t1) (6.25)

and so on

36

Let us check for example the second order term, by considering∫ t

t0

dt1

∫ t

t0

dt2T (HII (t1)H

II (t2)) =

∫ t

t0

dt1

∫ t1

t0

dt2HII (t1)H

II (t2)

+

∫ t

t0

dt1

∫ t

t1

dt2HII (t2)H

II (t1) (6.26)

The integral for the left-hand side is over the square (t0, t) × (t0, t). In thefirst integral of right-hand side is over the triangle white (t1 > t2) while thesecond term is integrated over the triangle (t2 > t1). However∫ t

t0

dt1

∫ t

t1

dt2HII (t2)H

II (t1) =

∫ t

t0

dt2

∫ t2

t0

dt1HII (t2)H

II (t1)

=

∫ t

t0

dt1

∫ t1

t0

dt2HII (t1)H

II (t2) (6.27)

so that∫ t

t0

dt1

∫ t1

t0

dt2T (HII (t1)H

II (t2)) = 2

∫ t

t0

dt1

∫ t1

t0

dt2HII (t1)H

II (t2) (6.28)

Dyson series can be written in a covariant form

S =∑n=0

(−i)n 1

n!

∫d4x1

∫d4x2 . . .

∫d4xnT (HI

I(x1)HII(x2) . . .HI

I(xn))

(6.29)The only source of non covariance of eq.(6.29) comes from the presence of theT ordering. However t1 − t2 > 0 is a property which remains true in everyLorentz frame when (x1 − x2)

2 is time-like while for space-like distances(x1 − x2)

2 < 0 [H(x1), H(x2)] = 0 and so the ordering is not relevant.

One usually defines the transition matrix T

Sfi = δfi − 2πiδ(Ef − Ei)Tfi (6.30)

and a perturbative series for S and T

Sfi = δfi + S(1)fi + S

(2)fi + . . . (6.31)

Tfi = T(1)fi + T

(2)fi + . . . (6.32)

37

The S matrix is unitaryS†S = I (6.33)

This is equivalent to the requirement of the conservation of probability.∑f

< i|S†|f >< f |S|i >= 1 (6.34)

The relation the with evolution operator in the Schrodinger representationis given by

UI(t, t0) = eiH0tUS(t, t0)e−iH0t0 (6.35)

In the application to scattering problems it is convenient sometime to turnon and off the interaction adiabatically to avoid problems with the oscillatorybehaviour at t→ ±∞. This is obtained by replacing the Hamiltonian with

HI → HIe−ε|t| (6.36)

so that the interaction acts for a time approximately of order 2/ε. Then atthe end of the calculations, after all the integrations have been performed,we take the limit ε→ 0.

To first order

−2πiδ(Ef − Ei)T(1)fi = S

(1)fi = −i lim

t0→−∞,t→∞

∫ t

t0

dt′ < f |HII |i > (6.37)

Recalling we have

−2πiδ(Ef − Ei)T(1)fi = −i lim

t0→−∞,t→∞

∫ t

t0

dt′ < f |eiH0t′HIe−iH0t′|i >

= −i limε→0+

∫ ∞

−∞dt′ < f |eiH0t′e−|ε|t′V e−iH0t′|i >

= −i limε→0+

∫ ∞

−∞dt′ei(Ef−Ei)t

′e−|ε|t′ < f |V |i >

= −i < f |V |i > limε→0+

[1

i(

1

Ef − Ei − iε)− 1

i(

1

Ef − Ei + iε)]

= − < f |V |i > [Pv1

Ef − Ei+ iπδ(Ef − Ei)

−Pv 1

Ef − Ei+ iπδ(Ef − Ei)]

= −2πiδ(Ef − Ei) < f |V |i > (6.38)

38

The same result can be obtained without the adiabatic approximation butworking with distributions.

In conclusionT

(1)fi =< f |V |i > (6.39)

where < f |V |i > is calculated in the Schrodinger representation (at t = 0).

The second order result is given by

T(2)fi =

∑n

< f |V |n >< n|V |i >Ei − En + iε

(6.40)

where |n > denotes the general intermediate state.

We have (defining τ = t1 − t2))

−2πiδ(Ef − Ei)T(2)fi = (−i)2 < f |

∫ ∞

−∞dt1

∫ t1

−∞dt2H

II (t1)H

II (t2)|i >

=

∫ ∞

−∞dt1

∫ t1

−∞dt2e

iEf t1 < f |H0e−iH0t1e+iH0t2HI |i > e−iEit2

=∑n

∫ ∞

−∞dt1

∫ t1

−∞dt2e

iEf t1e−iEit2e−iEn(t1−t2) < f |HI |n >< n|HI |i >

=∑n

∫ ∞

−∞dt1

∫ ∞

0

dτeiEf t1e−iEnτeiEi(τ−t1) < f |HI |n >< n|HI |i >

=∑n

< f |V |n >< n|V |i > 2πδ(Ef − Ei)

∫ ∞

0

dτei(Ei−En)τ

=∑n

< f |V |n >< n|V |i > 2π1

i(Ei − En + iε)(6.41)

where use has been made of the Fourier transforms in the distribution space(see Appendix B)

1√2π

∫ +∞

0

θ(x)eipxdx = − 1

i√2π

limε→0

1

p+ iε(6.42)

1√2π

∫ ∞

−∞dxeipx =

1√2πδ(p) (6.43)

For the proof with the adiabatic factor, see for example [5].

39

6.2 Fermi golden rule

Let us now compute the transition probability by considering the squaredmodulus of the S matrix element. Proceeding in this way one gets a diver-gence, generated by δ(0):

2πδ(Ef − Ei)2πδ(Ef − Ei)|Tfi|2 = 4π2δ(0)δ(Ef − Ei)|Tfi|2 (6.44)

It is possible to obtain a finite result, by considering a transition probabilityper time unit, assuming that the perturbation acts for a finite time T andthen taking the limit T → ∞. Let us consider the following representationof δ

2πδ(Ef − Ei) = limT→∞

∫ T/2

−T/2dtei(Ef−Ei)t (6.45)

So we have

limT→∞

Pfi(T )

T= lim

T→∞

1

T

∫ T/2

−T/2dtei(Ef−Ei)t lim

T→∞

∫ T/2

−T/2dtei(Ef−Ei)t|Tfi|2

= limT→∞

1

T

∫ T/2

−T/2dtei(Ef−Ei)t2πδ(Ef − Ei)|Tfi|2

= 2πδ(Ef − Ei)|Tfi|2 (6.46)

In general, when computing the rate of transition in a final state, one addsalso a factor taking into account the density of final states or phase spacedφf

dwfi = 2πδ(Ef − Ei)|Tfi|2dφf (6.47)

For example, when in the final state we have a particle with momentum k

dφf = V 3 d3kf

(2π)3(6.48)

To compute the cross section, defined as the transition rate in a groupof final states for one scattering center and unit incident flux, we need tocalculate

dσ =dwfiΦ

(6.49)

where Φ is the flux of incoming particles.

40

7 Radiation processes of first order: emission

and absorption of a single photon

We are now ready to compute the emission and the absorption of a photonby an atom. Let us first consider the decay of an atom which emits a photonof momentum k and polarization α. The initial state is

|i >≡ |A; . . . nk,α . . . >≡ |A > ⊗| . . . nk,α . . . > (7.1)

and the final state

|f >≡ |A′; . . . nk,α + 1 . . . >≡ |A′ > ⊗| . . . nk,α + 1 . . . > (7.2)

where |A(A′) > denotes the initial (final) atom state. We have to computethe matrix element

Vfi =< A′; . . . nk,α + 1 . . . |V1|A; . . . nk,α . . . > (7.3)

where V1 is given in Eq.(5.12). In the coordinate space the initial (final) stateis described by the wave function

ψA(A′)(ξr) =< ξ1, · · · ξN |A(A′) > (7.4)

To first order in the perturbation theory we have ( neglecting proton con-tribution with respect to electron one and denote the electron masses bym)

T(1)fi = − e

m

∑r

< A′; . . . nk,α + 1 . . . |pr ·A(0, ξr)|A; . . . nk,α . . . >

= − e

m

∑r

∑k′,α′

1√2V ωk′

< A′; . . . nk,α + 1 . . . |pr · εα′

k′ [aα′

k′ exp (ik′ · ξr) + h.c.]

|A; . . . nk,α . . . >

= − e

m

∑r

∑k′,α′

1√2V ωk′

< A′|pr · εα′

k′ exp (−ik′ · ξr)|A > δα,α′δk,k′√nk′,α′ + 1

= − e

m

∑r

√nk,α + 1√2V ωk

< A′|pr · εαk exp (−ik · ξr)|A >

= − e

m

∑r

√nk,α + 1√2V ωk

∫d3ξrψ

∗A′(ξr)ε

αk · (−i∇r)(exp (−ik · ξr)ψA(ξr))

(7.5)

41

If the transitions are in the visible region, k−1 ∼ 1000 Angstrom, thenkξr << 1 since ξr ∼ 1 Angstrom. So, in the dipole approximation, exp (−ik · ξr) ∼1,

T(1)fi = − e

m

√nk,α + 1

2V ωk

∑r

< A′|pr · εαk)|A >

= −e√nk,α + 1

2V ωk

∑r

< A′|vr · εαk)|A > (7.6)

where vr is the velocity operator.

Let us now compute the probability of emitting photons of momentumin the interval k,k+ dk and polarization α

dwfi = 2πδ(Ef − Ei)|T (1)fi |

2V d3k

(2π)3

= 2πe2

2V ωk(nk,α + 1)|

∑r

< A′|vr · εαk(exp (−ik · ξr))|A > |2δ(Ef − Ei)V d3k

(2π)3

=d3k

(2π)2(nk,α + 1)

e2

2ωk|∑r

< A′|vr · εαk(exp (−ik · ξr))|A > |2δ(EA′ + ωk − EA)

=αem2π

(nk,α + 1)ωkdωkdΩ|∑r

< A′||vr · εαk(exp (−ik · ξr))|A > |2δ(EA′ + ωk − EA)

(7.7)

We can now integrate over ωk obtaining the probability of emitting a photonin the solid angle dΩ

dwfidΩ

=αem2π

ωk(nk,α + 1)|∑r

< A′|vr · εαk(exp (−ik · ξr))|A > |2 (7.8)

where now ωk = EA − EA′ . In the dipole approximation

|∑r

< A′|vr·εαk(exp (−ik · ξr))|A > |2 ∼ |∑r

< A′|vr·εαk|A > |2 ∼ | < A′|De·εαk|A > |2

(7.9)where we have introduced the dipole operator

D = e∑r

ξr (7.10)

42

UsingD = i[Hatom,D] (7.11)

we getDA′A = i < A′|[Hatom,D]|A >= i(EA′ − EA)DA′A (7.12)

thereforedwfidΩ

=αem2π

ω3k(nk,α + 1)|1

eDA′A · εαk|2 (7.13)

with1

eDA′A =

∫(Πrd

3ξr)ψ∗A′(ξr)

∑r

ξrψA(ξr) (7.14)

The result is zero also when there is no initial radiation field, nk,α = 0. This isthe new result which emerges at the quantum level, the so-called spontaneousemission.

When nk,α = 0, and we do not observe the polarization of the photon, wecan perform some more analytical step:∑

α

dwfidΩ

=αem2π

ω3k

∑α

|1eDA′A · εαk|2 (7.15)

Since∑α

|1eDA′A·εαk|2 =

1

e2

∑α

Di∗A′AD

jA′Aε

αiεαj =1

e2Di∗A′AD

jA′A(δij−

kikjk2

) =1

e2|D⊥

A′A|2

(7.16)In conclusion ∑

α

dwfidΩ

=αem2π

ω3k

1

e2|D⊥

A′A|2 (7.17)

By integrating over all the solid angle, we obtain the life-time τ for thetransition A→ A′ which is defined by

1

τAA′=

∫dΩ

∑α

dwfidΩ

=αem2π

ω3k

1

e2|DA′A|2(4π−

4

3π) =

4

3ω3k

1

e2|DA′A|2 (7.18)

To get the total life time one has to sum over all the possible states in whichthe atomic state can decay:

1

τA=

∑A′

1

τAA′(7.19)

43

In the cgs system the eq. (7.18) becomes

1

τ=

4

3ω3k

1

c2e2|DA′A|2 (7.20)

To get an extimate of the life time for the 2p to 1s transition for the hydrogenatom, assume ωk ∼ 1016 sec−1, D/e ∼ 0.5× 10−8 cm, then we get τ ∼ 10−8

sec.

8 Interaction of the light with the matter

Let us now consider the scattering of the light (photons) by atomic electrons.We consider the Hamiltonian given in eq. (5.9). In particular we will computethe scattering of a photon of momentum k1 and polarization α1 by an electronbound in a given atom. For simplicity we consider just an atom with oneelectron. The result can be easily generalized to N electrons. The initialstate is

|A > ⊗|k1, α1 > (8.1)

while the final state is|A′ > ⊗|k2, α2 > (8.2)

where k2 and α2 are the momentum and the polarization of the final statephoton, |A(A′) > denote the initial (final) electron state. In conclusion inthe process the number of photons does not change ∆n = 0. Since theHamiltonian V1 can describe only ∆n = ±1 process, we have to go to thenext order in the expansion, that is to order (e2). The Hamiltonian V2 candescribe ∆n = 0 process, since

V2 ∼ (aαk + aα†k )(aα′

k′ + aα′†

k′ ) (8.3)

The general amplitude to order e2 is given by

T(2)fi = (V2)fi +

∑n

(V1)fn(V1)niEi − En + iε

(8.4)

where the sum is over all possible intermediate states |n >≡ |N > ⊗|...nk,α... >,

Ei = EA + ωk1 (8.5)

44

En = EN + ... denotes the energy of the intermediate state. Let us firstconsider the case when the energy of the incoming photon is much largerthan the level splitting of the atoms. In this case one can neglect the secondterm of eq.(8.4) and one recovers the classical result of the scattering of thelight by a free electron (Thomson5 scattering). Substituting in V2 we get

V2 =e2

2m

∑k,α

∑k′,α′

εαk · εα′

k′1√

2ωkV√2ωk′V

(aαkeik·ξ +h.c.)(aα

′

k′eik′·ξ +h.c.) (8.6)

The terms contributing to the final result are aαkaα′†k′ and aα†k a

α′

k′ . The finalresult is

T(2)fi = (V2)fi =

e2

mεα1k1

· εα2k2

1

2V√ωk1ωk2

< A′|ei(k1−k2)·ξ|A >

=e2

mεα1k1

· εα2k2

1

2V√ωk1ωk2

∫d3ξψ∗

A′(ξ)ei(k1−k2)·ξψA(ξ) (8.7)

where we have introduced the wave function of the electron in the stateA(A′), ψA(A′)(ξ). If the dipole approximation can be used, we get

T(2)fi = (V2)fi =

e2

mεα1k1

· εα2k2

1

2V√ωk1ωk2

δAA′ (8.8)

Using the Fermi golden rule we get the probability of scattering per unit time

dw = 2πδ(ωk1 − ωk2)e4

4ωk12V 2m2

(εα1k1

· εα2k2)2

V

(2π)3d3k2 (8.9)

Finally the cross section is obtained by dividing by the incoming flux Φ =ρv = 1/V , being the photon velocity equal to one in natural units,

dσ =dw

Φ= δ(ωk1 − ωk2)(

e2

4πmωk1εα1k1

· εα2k2)2k22dk2dΩk2 (8.10)

By integrating over ωk2 = k2∫dk2dσ = r20(ε

α1k1

· εα2k2)2dΩk2 (8.11)

5J.J. Thomson, 1856-1940, Nobel prize in Physics in 1906 for the discovery of theelectron

45

where r0 is the classical radius of the electron

r0 =e2

4πm∼ 2.8fm = 2.8× 10−13cm (8.12)

If we do not know the initial polarization and we do not measure the finalpolarization, we average over the initial polarization and we sum over thefinal ones

1

2

∑α1,α2

dσ =r202(1 + cos θ2)dφd cos θ (8.13)

where θ is the angle between k1 and k2. In fact we have

1

2

∑α1,α2

εα1k1

iεα2k2

iεα1k1

jεα2k2

j =1

2

∑α1

εα1k1

iεα1k1

j(δij −ki1k

j1

k21)

=1

2(δij −

ki2kj2

k22)(δij −

ki1kj1

k21)

=1

2[3− 1− 1 + (k1 · k2)

2]

=1

2[1 + (k1 · k2)

2] (8.14)

By integrating over the angles, we obtain the total Thomson cross section

σ =8π

3r20 (8.15)

Using the numerical value of r0 we get σ = 6.6× 10−25cm2.

We can now proceed to compute the cross section in the general casederiving the so called Kramers Heisenberg formula (1925). The possibleintermediate states contributing to the second order part of the amplitudesare

|N > ⊗|0 >, |N > ⊗|k1, α1; k2, α2 > (8.16)

that is |N > times the vacuum states and the two photon state. The resultfor the amplitude is

T(2)fi = (V2)fi +

∑N

[(V 0

1 )A′N(V01 )NA

EA + ωk1 − EN + iε+

(V II1 )A′N(V

II1 )NA

EA − ωk2 − EN + iε] (8.17)

46

where

(V 01 )A′N =

e

m

1√2ωk2V

< A′|p · εα2k2e−ik2·ξ|N > (8.18)

(V 01 )NA =

e

m

1√2ωk1V

< N |p · εα1k1eik1·ξ|A > (8.19)

(V II1 )A′N =

e

m

1√2ωk1V

< A′|p · εα1k1eik1·ξ|N > (8.20)

(V II1 )NA =

e

m

1√2ωk2V

< N |p · εα2k2e−ik2·ξ|A > (8.21)

and (V2)fi is given in eq.(8.7). In the dipole approximation we get

T(2)fi =

1

2V√ωk1ωk2

fA′A (8.22)

with

fA′A =e2

mεα1k1

· εα2k2δAA′+ e2

m2

∑N

[< A′|p · εα2

k2|N >< N |p · εα1

k1|A >

EA + ωk1 − EN + iε

+< A′|p · εα1

k1|N >< N |p · εα2

k2|A >

EA − ωk2 − EN + iε] (8.23)

To obtain the cross section we divide by the flux:

dσ

dΩ= 2πδ(EA + ωk1 − EA′ − ωk2)

1

4V 2ωk1ωk2|fA′A|2k22dk2

V

(2π3)

11V

(8.24)

By integrating over k2, we obtain∫dσ

dΩ=ωk2ωk1

|fA′A

4π|2 (8.25)

The case A′ = A ωk1 = ωk2 correspond to the elastic case (Rayleigh scatter-ing) while the inelastic case, A′ 6= A ωk1 6= ωk2 , corresponds to the Raman6

effect.

Note that when the energy of the initial photon ωk1 = EN − EA, theamplitude as given in eq.(8.23) diverges and consequently the cross section

6C.V. Raman (1888-1970), Nobel prize in Physics in 1930

47

becomes infinite. Since the cross section is a measurable quantity, this meansthat the calculation to second order becomes inadequate and higher ordersbecome relevant.

What is happening is that we have assumed the intermediate states Nas stable, neglecting their life time due to the instability for the spontaneousemission. If the probability of finding an electron in a generic energy levelEN decreases for spontaneous emission as

exp (− t

τN) = exp (−ΓN t) (8.26)

where τN (ΓN) is the life time (width) of the state N , the correspondingamplitude behaves as

exp (−ΓN t/2) (8.27)

This is equivalent to a time evolution of the state as

exp [−i(EN − iΓN2)t] (8.28)

In conclusion in presence of a resonant process, when ωk1 ∼ EN −EA we canreplace

1

EA + ωk1 − EN→ 1

EA + ωk1 − EN + iΓN

2

(8.29)

so that the cross section, close to the resonance, assumes the classic Lorentzform

dσ

dΩ∼ |C|2

|EA + ωk1 − EN + iΓN

2|2

=|C|2

(EA + ωk1 − EN)2 +Γ2N

4

(8.30)

In order to compute the total width, we start considering the total shiftin the energy of the bound state to order O(e2):

∆En =< n|HI |n > +∑m6=n

< n|HI |m >< n|HI |m >

En − Em(8.31)

where|n >= |N > ⊗|0 > (8.32)

48

is a pure atom state, no photons are present. The interaction Hamiltonian is

HI = V1 + V2 (8.33)

with

V1 = − e

mp ·A(ξ, 0), V2 =

e2

2m(A(ξ, 0))2 (8.34)

Assuming HI normal ordered

< n|HI |n >=< n|V2|n >= 0 (8.35)

and only the second term of eq. (8.36) is different from zero

∆En =∑m6=n

< n|V1|m >< m|V1|n >En − Em

(8.36)

Therefore since |n >= |N > ⊗|0 > the only intermediate state allowed is

|m >= |M > ⊗|1k,α > (8.37)

and the result is given by the amplitude for the spontaneous emission

(< N |⊗ < 0|)V1(|M > ⊗|1k,α >) = T spont.emiss.NM (8.38)

where we can use the result of the spontaneous emission (7.8) for nk,α = 0for one electron or

T spont.emiss.MN = − e

m

1√2V ωk

< N |p · εαk exp (−ik · ξ)|M >

= − e

m

1√2V ωk

|tMN |2 (8.39)

So we get

∆En =e2

2m2V

∑k,α

∑M

1

ωk

|tMN |2

EN − EM − ωk

→ e2

2m2

∫d3k

(2π)3

∑M

1

ωk

|tMN |2

EN − EM − ωk

=α

m2

1

(2π)2

∫dωkdΩωk

∑M

|tMN |2

EN − EM − ωk

(8.40)

49

On the other hand using (7.8) for spontaneous emission for atoms with asingle electron, we have

dwNMdΩ

=α

2πm2ωk| < N |p · εαk(exp (−ik · ξ)|M > |2

=α

2πm2ωk|tMN |2

(8.41)

where ωk = EN − EM .

In conclusion we can rewrite ∆EN as

∆En =1

2π

∑M

∫ ∞

0

dωk1

EN − EM − ωk

dwNMdΩ

(8.42)

where dwNM

dΩis the probability of spontaneous emission of a photon of fre-

quency ωk which assumes any value from zero to infinity.

The integral (8.44) has a pole for ωk = EN − EM . Let us replace thedenominator with EN − EM − ωk + iε, so we have to compute

∆En =1

2π

∑M

∫ ∞

0

dωk1

EN − EM − ωk + iε

dwNMdΩ

(8.43)

Using (B.6) we have

∆En =1

2π

∑M

[Pv

∫ ∞

0

dωk1

EN − EM − ωk−iπ

∫ ∞

0

dωkδ(EN−EM−ωk)]dwNMdΩ

(8.44)

By taking the imaginary part

Im[∆En] = − 1

2π

∑M

∫ ∞

0

dωk[πδ(EN − EM − ωk)]dwNMdΩ

= −1

2

∑M

dwNMdΩ

|ωk=EN−EM

= −1

2ΓN (8.45)

50

where ΓN is the total width of the state N due to spontaneous emission.Therefore the time evolution is now

ψN(t) = e−i(EN+Re[∆EN ]− i2ΓN )tψN(0) = e−i(EN+Re[∆EN ])te−

12ΓN tψN(0) (8.46)

and the probability decreases as

|ψN(t)|2 = e−ΓN t|ψN(0)|2 (8.47)

The real part of ∆EN gives the shift in the energy of the bound state corre-sponding to the Lamb shift. The integral

Re[∆En] =1

2π

∑M

Pv

∫ ∞

0

dωk[1

EN − EM − ωk]dwNMdΩ

(8.48)

is divergent and a special procedure of renormalization of the mass of theelectron is necessary [18, 17].

9 Cherenkov effect

In 1934, while Pavel Cherenkov (1904-1990) was studying, under the super-vision of Sergei Vavilov at the Physical Institute of the USSR Academy ofScience, the luminescence of liquids of uranyl salt under irradiation of gammarays from radium, he discovered a new bleu glow. Vavilov suggested that theeffect could be due to bremstrahlung of electrons that were knocked out bythe gamma rays of the radium. However the correct explanation, which isnot bremstrahlung, was given by Tamm and Frank7 (1937): they consideredthe field of a point like charged particle moving in a medium uniformly andrectilinearly and show that if the velocity v of the particle is higher than thevelocity of the light in the medium c/n, the particle emits a radiation in acone of angle θ such that

cos θ =c

nv(9.1)

Let us first consider when a free charged particle of momentum p, travelingin a medium of index n, can emit a photon of momentum k. From theconservation of the momentum we get

pµ = (p′ + k)µ (9.2)

7I.M. Franck, I.E. Tamm, Nobel prize in Physics in 1958 with Cherenkov

51

or(p′)2 = m2 = (p− k)2 = m2 − 2pµkµ + ω2

k − k2 (9.3)

We conclude that

2(p0ωk − p · k) = ω2k − k2 = (

1

n2− 1)k2 (9.4)

Then

p · k = Ek

n+n2 − 1

2n2k2 (9.5)

or

cos θ =E

p

1

n+n2 − 1

2n2

k

p

=1

nv+n2 − 1

2n

ω

mγv(9.6)

The exact relativistic formula for the Cherenkov angle is

cos θ =1

nv

[1 +

n2 − 1

2mω√1− v2

](9.7)

where we have used E = mγ, p = mγv.

Since the Cherenkov light in the water is in the visible light correspondingto 400-700 nm∼ or 10−1 eV−1 the ratio k/m ∼ 10−4, therefore the angle isapproximately

cos θ =1

nv(9.8)

This result implies1

nv≤ 1 (9.9)

or v ≥ 1/n. So in the vacuum this process is forbidden, since v cannot begreater the 1, (the light velocity c = 1 in natural unit). However in a mediumv can be greater than 1/n and the process is allowed.

In conclusion a charged particle, which travels at a velocity that exceedsc/n, can emit a photon. The refractive index of the water between 400-700nm is in the range 1.33-1.34. So the critical velocity is v ∼ 0.75.

52

This effect has been used to build detectors for charged particles. Todaythe largest Cherenkov detector is the Super-Kamiokande detector, 40 m x40 m approx, 50000 tons of water, 11200 photomultipliers, located in Japan.This experiment has discovered neutrino oscillations. The Cherenkov lightis emitted in a cone shape to the direction of a charged particle. The photo-multiplier tubes of the tank detect this Cherenkov light and give informationof the quantity of the detected light and the timing of the detection. Theygive information also on the energy, direction, interaction point and type ofthe charged particle.

We will derive the Cherenkov emission of a charged particle by consid-ering the quantized electromagnetic field in a medium. For the total energyassociated to the electromagnetic field see [6, 7]. In order to get the standardform for the Hamiltonian in terms of creation and annihilation operators weneed to include the refraction index in the expansion [6, 7]

A(x) =∑k

∑α=1,2

1

n√2V ωk

εαk[aαke

−ikx + h.c.], ki = ni2π

L(9.10)

where now k0 ≡ ωk = k/n and n = n(ωk). To calculate the probability thatan electron emits Cherenkov radiation, for simplicity we will consider thenon relativistic approximation for the electron. The initial state is

|i〉 = |p〉 ⊗ |0〉 (9.11)

where |p〉 represent a non relativistic electron state of momentum p and |0〉the state with no photons. The final state is

|f〉 = |p′〉 ⊗ |1k,α〉 (9.12)

where |1k,α〉 represent the Cherenkov photon. Let us compute

Vfi = 〈f |V |i〉 = 〈p′| ⊗ 〈1k,α|(−ep

m·∑k′

∑α′=1,2

1

n√2V ωk′

εα′

k′ [aα′

k′eik′·x + h.c.])|p〉 ⊗ |0〉

= 〈p′|(−e pm

· εαk1

n√2V ωk

e−ik·x)|p〉

= − ep′·εαknm

√2V ωk

〈p′|e−ik·x)|p〉

= − ep′·εαknm

√2V ωk

δp′−p+k,0

= − ep·εαknm

√2V ωk

δp′+k−p,0 (9.13)

53

where we have used the non relativistic wave functions

< x|p >= 1√Veip·x (9.14)

Then one can compute the probability that an electron of momentum p emitsa photon of momentum k. Passing to the continuum, we get

dw = 2πδ(Ef − Ei)|Vfi|2V

(2π)3d3p′

V

(2π)3d3k

= 2πδ(Ef − Ei)(ep·εαk

nm√2V ωk

)2(2π)3

Vδ3(p′ + k − p)

V

(2π)3d3p′

V

(2π)3d3k

=πe2

n2ωk| pm·εαk|2

d3p′

(2π)3d3kδ3(p′ + k − p)δ(Ep′ + ωk − Ep) (9.15)

Integrating over d3p′∫p′dw =

πe2

n2ωk| pm·εαk|2

1

(2π)3d3kδ(Ep′ + ωk − Ep)

∼ πe2

n2ωk| pm·εαk|2

1

(2π)3d3kδ(v · k− ωk)

=e2

8π2| pm·εαk|2nωkdωkd cos θdφδ(v · k− ωk) (9.16)

where we have used

Ep′ = Ep−k ∼ Ep −∂E

∂p· k (9.17)

However

δ(v · k− ωk) =1

vkδ(cos θ − 1

nv) (9.18)

Therefore we get∫p′dw =

e2

8π2|v·εαk|2nωkdωkd cos θdφ

1

vkδ(cos θ − 1

nv)

=e2

8π2|v·εαk|2dωkdφd cos θ

1

vδ(cos θ − 1

nv) (9.19)

Summing over the polarization and finally integrating over φ and θ∫φ,cos θ

∑α

∫p′dw =

∫φ,cos θ

∑α

e2

8π2vivjεαiεαj

1

vδ(cos θ − 1

nv)dωk

54

=e2

4πv(1− 1

n2v2)dωk

= αv(1− 1

n2v2)dωk (9.20)

The energy loss of the electron per unit lenght of the trajectory is

dE

dx=

1

v

dE

dt=

∫ ωmax

0

1

vαv(1− 1

n2v2)ωkdωk = α

∫ ωmax

0

(1− 1

n2v2)ωkdωk

(9.21)The integral is cutoff by ωmax,

ω ≤ ωmax ∼(nv − 1)2m

n2 − 1

γ

v(9.22)

as we obtain by requiring cos θ ≤ 1 from eq.(9.7). The number of photonsemitted per unit lenght and unit of wave lenght is

dN

dxdλ= α(1− 1

n2v2)2π

λ2(9.23)

This explains the blue color of the Cherenkov radiation, since the intensitygrows at short wave lenghts, blue and violet.

10 The Dirac field

10.1 The Dirac equation: classical theory

Let us now consider the classical theory of the Dirac equation (1928), whichdescribes relativistic particles with spin 1

2. Let us first recall the Dirac equa-

tion in non covariant form:

i∂ψ(x)

∂t= [α · (−i∇) + βm]ψ(x) (10.1)

where αi, β are hermitian 4× 4 matrices, satisfying

β2 = 1, [β, αi]+ = 0, [αi, αj]+ = δij (10.2)

55

In the Dirac-Pauli representation

β =

(1 00 −1

), α =

(0 σσ 0

)(10.3)

The equation can be rewritten in the form

(iγµ∂µ −m)ψ(x) = 0 (10.4)

whereγ0 = β, γi = βαi (10.5)

are Dirac 4× 4 matrices satisfying

[γµ, γν ]+ = 2gµν (10.6)

and㵆 = γ0γµγ0 (10.7)

In the Dirac-Pauli representation

γ0 =

(1 00 −1

), γ =

(0 σ

−σ 0

)(10.8)

Therefore ψ(x) is a four component wave function. Usually the componentsare indicated by ψα(x), α = 1, 2, 3, 4.

10.2 Lorentz and parity transformation

Let us now assume that the Dirac equation be valid in any inertial frame.Then in the S ′ system, where x′ = Λx, assuming the relativity principle,Dirac equation must have the same form

(iγµ′∂′µ −m)ψ′(x′) = 0 (10.9)

In order to reproduce the Klein-Gordon condition the matrices γµ′ mustsatisfy the same algebra as the matrices γµ. By requiring also the samehermiticity condition (10.7), neglecting a unitary transformation, we canalways identify γµ′ ≡ γµ.

56

Assuming that

ψ′(x′) = S(Λ)ψ(x) = S(Λ)ψ(Λ−1x′) (10.10)

we can prove that, if for an infinitesimal transformation we write Λ as

Λµν ∼ gµν + εµν (10.11)

then

S(Λ) ∼ 1− i

4σµνε

µν (10.12)

with

σµν =i

2[γµ, γν ] (10.13)

with σµν = −σνµ.

In fact from the Dirac equation in the S system, we get

0 = (iγµ∂µ −m))S−1(Λ)ψ′(x′)

=

(iγµ

∂x′ν

∂xµ∂′ν −m)

)S−1(Λ)ψ′(x′)

=(iγµΛνµ∂

′ν −m)

)S−1(Λ)ψ′(x′)

(10.14)

where ∂′µ = ∂∂x′µ

. By multiplying by S(Λ) to the left, we get

(iS(Λ)γµS−1(Λ)Λνµ∂

′ν −m)

)ψ′(x′) = 0 (10.15)

Comparing with the Dirac equation in the S ′ system we get

S(Λ)γµS−1(Λ)Λνµ = γν (10.16)

orS−1(Λ)γνS(Λ) = Λνµγ

µ (10.17)

For the infinitesimal transformation (10.11) and assuming (10.12) to firstorder in ερλ we get

[σρλ, γν ] = −2i(gρνγλ − gλνγρ) (10.18)

57

The unique solution to eq.(10.18) is given by

σρλ =i

2[γρ, γλ] (10.19)

For a finite transformation the solution for S(Λ) is obtained by exponen-tiation of eq.(10.12):

S(Λ) = exp(− i

4σµνε

µν) (10.20)

S(Λ) is the representation of the Lorentz transformation Λ on the space ofthe wave functions ψ. The six matrices σµν are the generators of the Lorentztransformations. In particular σ0i are the three generators of the Lorentzboosts and σij are the three generators of the 3D-rotations. Using the DiracPauli representation for the gamma matrices

γ0 =

(1 00 −1

), γk =

(0 σk

−σk 0

)(10.21)

we have

σij = εijk

(σk 00 σk

)(10.22)

Let us now prove that ψψ ≡ ψ†γ0ψ is a scalar under Lorentz transforma-tion. In fact from eq.(10.10)

ψ† → ψ†S(Λ)† (10.23)

andψ → ψ†S(Λ)†γ0 = ψγ0S(Λ)†γ0 = ψS−1(Λ) (10.24)

where in the last term we have used the property (10.7). Therefore

ψψ → ψS−1(Λ)S(Λ)ψ = ψψ (10.25)

In analogous way we can prove that ψγµψ transforms as a fourvector. Infact

ψγµψ → ψS−1(Λ)γµS(Λ)ψ = Λµνψγνψ (10.26)

where use has been made of (10.17).

58

Finally we can also prove that ψσµνψ transform as a tensor:

ψσµνψ → ΛµρΛµλψσ

ρλψ (10.27)

Let us now study the parity transformation

x → −x (10.28)

which is assumed to be a symmetry of the Dirac equation. The parity trans-formation acts on the fourdimensional space as

x′µ = P µν x

ν , P µν =

1

−1−1

−1

(10.29)

The transformation of the spinor field under the parity transformation isobtained by requiring the invariance of the Dirac equation under parity andit is given by

ψ(x) → ψ′(x′) = S(P )ψ(x) (10.30)

with S(P ) such that

S−1(P )γ0S(P ) = γ0, S−1(P )γiS(P ) = −γi (10.31)

In this way we obtain:

(iγ0∂0 − iγi∂i −m)ψ(x) = 0

→P (iγ0∂0 + iγi∂′i −m)S−1(P )ψ′(x′) = 0 (10.32)

By multiplying by S(P ) and using (10.31) we obtain

(iγ0∂0 − iγi∂′i −m)ψ′(x′) = 0 (10.33)

The solution of eq.(10.31) is given by

S(P ) = γ0ηP (10.34)

with |ηP | = 1.

59

Using the matrix γ5 defined as8

γ5 = iγ0γ1γ2γ3 (10.35)

with the properties

γ†5 = γ5, γ25 = 1, [γ5, γµ]+ = 0 (10.36)

we can consider the expressionψγ5ψ (10.37)

Under the Lorentz transformation

ψγ5ψ → ψγ5ψ (10.38)

However under the parity transformation

ψγ5ψ → −ψγ5ψ (10.39)

In conclusion while ψψ is a scalar, the ψγ5ψ bilinear is a pseudo-scalar.In analogous way one can prove that

ψγ5γµψ (10.40)

is an axial-four-vector.

Note that the 16 matrices (1, γ5, γµ, γ5γµ, σµν) are a basis for the 4 by 4hermitian matrices.

10.3 Wave plane solutions of the Dirac equation

Let us now look for solutions of the Dirac equation of the form

ψ(x) = e−ipxu(p), positive energy

ψ(x) = eipxv(p), negative energy (10.41)

8An alternative form for γ5 is γ5 = i24εµνρσγ

µγνγργσ. This form is useful to show thatψγ5ψ trasforms as a pseudoscalar, using S−1(Λ)γ5S(Λ) = (detΛ)γ5. In the Dirac-Pauli

representation γ5 =

(0 11 0

)

60

wherepx ≡ p0t− p · x ≡ Et− p · x (10.42)

withE =

√p2 +m2 (10.43)

Dirac equation implies:

(p−m)u(p) = 0, (p+m)v(p) = 0 (10.44)

Going into the rest frame pµ = (m,0), eqs.(10.44) become

(γ0 − 1)u(0) = 0, (γ0 + 1)v(0) = 0 (10.45)

In the Dirac-Pauli representation, where γ0 is given by (10.21) the solutionsare

u1(0) =

1000

, u2(0) =

0100

, v1(0) =

0010

, v2(0) =

0001

,

(10.46)Note that we have two positive energy and two negative energy independentsolutions. u1,2 and v1,2 are eigenvectors of the generator of rotations alongthe z axis

σ122

=1

2

(σ3 00 σ3

)(10.47)

corresponding to eigenvalues ±1/2. The solutions to a generic frame canbe obtained by boosting these solutions from the rest frame with a Lorentztransformation with velocity v = p/E (see [2]). However it is simpler toobserve that

(p−m)(p+m) = γµpµγνpν −m2

=1

2[γµ, γν ]+pµpν −m2 = p2 −m2 = 0 (10.48)

where use has been made of (10.6). Therefore we have

ur(p) ∼ (p+m)ur(0), vr(p) ∼ (p−m)vr(0), r = 1, 2 (10.49)

By requiring the normalization

ur(p)us(p) = δrs, vr(p)vs(p) = −δrs (10.50)

61

we get (see Appendix E.1)

ur(p) =p+m√

2m(E +m)ur(0), vr(p) =

−p+m√2m(E +m)

vr(0), r = 1, 2

(10.51)We have also the orthogonality condition

ur(p)vs(p) = 0, vr(p)us(p) = 0 (10.52)

and the completeness condition∑r

[urα(p)urβ(p)− vrα(p)vrβ(p)] = δαβ, α, β = 1, 2, 3, 4 (10.53)

Now the complete set

1√Ve−ipxur(p),

1√Veipxvr(p) (10.54)

can be used to expand the general solution of the Dirac equation

ψ(x) =∑rp

√m

V E[br(p)ur(p)e

−ipx + d∗r(p)vr(p)eipx] (10.55)

with br(p), d∗r(p) complex functions that after quantization of the spinor field

ψ become operators. Notice that we have also (see Appendix E.1)

u†r(p)us(p) =E

mδrs = v†r(p)vs(p) (10.56)

u†(p)v(−p) = v†(p)u(−p) = 0 (10.57)

10.4 Lagrangian of the Dirac field

The Dirac equation given in eq.(10.4) can be derived from the Lagrangiandensity

L = ψ(x)[i∂ −m]ψ(x) (10.58)

Varying the action with respect to ψ, we obtain the Dirac equation. Assum-ing the mass as fundamental dimension, the dimension of the Dirac field ar

62

M3/2 so that the action is dimensionless. The interaction of the relativis-tic electron with the electromagnetic field is obtained by using the minimalsubstitution

E → E − eA0, p → p− eA (10.59)

or in the covariant formpµ → pµ − eAµ (10.60)

andi∂µ → i∂µ − eAµ (10.61)

where Aµ is the four potential of the electromagnetic field. The Lagrangiandescribing the Dirac field interacting with the electromagnetic field is there-fore

L′ = ψ(x)[i∂ − eA−m]ψ(x) (10.62)

and the corresponding Dirac equation is

[i∂ − eA−m]ψ(x) = 0 (10.63)

10.5 Non relativistic limit of the Dirac equation

Let us now consider the non relativistic limit of the Dirac equation in anexternal field. Starting from (10.63) and multiplying by γ0 we get

i∂ψ(x)

∂t= [α · (−i∇− eA) + βm+ A0]ψ(x) (10.64)

Let us now look for solutions with positive energy of the following form

ψ(x) = exp (−iEt)(φ(x)χ(x)

)(10.65)

where φ, χ are two component spinors depending on x. We assume

E = m+ T (10.66)

with T << m. We obtain

(m+ T )

(φχ

)=

((m+ eA0)φ+ σ · πχσ · πφ+ (−m+ eA0)χ

)(10.67)

63

with π = −i∇− eA or

(eA0 − T )φ+ σ · πχ = 0

σ · πφ+ (−2m+ eA0 − T )χ = 0 (10.68)

By considering A0, T << m from the second equation of (10.68) we get

χ =σ · πφ2m

(10.69)

By substituting (10.69) in the first equation of (10.68) we obtain

Tφ = (1

2mσ · π · σ · π + eA0)φ (10.70)

Using

σ · π · σ · π = σiσjπiπj =1

2([σi, σj] + [σi, σj]+)πiπj = iεijkσkπiπj + π2

=i

2εijkσk[πi, πj] + π2

= −1

2εijkσk(∂iAj − ∂jAi) + π2

= −εijkσk∂iAj + π2

= −σ ·B + π2 (10.71)

we get

Tφ = (1

2mπ2 − 1

2mσ ·B + eA0)φ

= (1

2mπ2 − 1

2m2S ·B + eA0)φ

= (1

2mπ2 − µ ·B + eA0)φ (10.72)

where S denotes the electron spin and µ the magnetic momentum of theelectron

µ =1

2m2S (10.73)

In conclusion Dirac equation predicts the magnetic field the gyromagneticfactor of the electron

ge = 2 (10.74)

64

Its experimental value is different from 2 at the per mil level (Particle DataGroup, PDG)

ge − 2

2= 1159.65218076± 0.00000027× 10−6 (10.75)

By studying the next order in the non relativistic expansion of the Diracequation and assuming for A0 the Coulomb potential for the Hydrogen atom,one can get the fine structure terms, see [17]: the relativistic correction

− p4

8m3(10.76)

the Darwin terme

8m2∇2A0 (10.77)

and the spin-orbit terme

2m2

1

r

dφ

drS ·L (10.78)

10.6 Quantization of the Dirac field

The Dirac equation given in eq.(10.4) can be derived from the Lagrangiandensity (10.58) by minimizing the action with respect to ψ. The conjugatemomenta are given by

πα(x) =∂L∂ψα

= iψ†α, πα(x) =

∂L∂ ˙ψα

= 0 (10.79)

The Hamiltonian is given by

H =

∫d3xH (10.80)

with

H = π(x)ψ(x) + π(x) ˙ψ(x)− L = ψ(x)[−iγj∂j +m]ψ(x) (10.81)

When computing the Hamiltonian, using the field expansion (10.55), wehave L = 0 and therefore the Hamiltonian is simply given by

H =

∫d3xψ†(x)ψ(x) (10.82)

65

Therefore substituting in the Hamiltonian the expansion (10.55) we obtain

H =

∫d3x

∑rp

√m

V E[b†r(p)u

†r(p)e

ipx + dr(p)v†r(p)e

−ipx]

∑sp′

E ′√

m

V E ′ [bs(p′)us(p

′)e−ip′x − d†s(p

′)vs(p′)eip

′x]

=m

V

∑rp

∑sp′

1√EE ′

[(b†r(p)bs(p

′)u†r(p)us(p′)

+ dr(p)d†s(p

′)v†r(p)vs(p′))δp,p′

+(dr(p)bs(p

′)v†r(p)us(p′)− b†r(p)d

†s(p

′)u†r(p)vs(p′))δp,−p′

](10.83)

and using the orthogonality properties of the spinors, eq.(10.56) and eq.(10.57),

H =∑pr

E[b†r(p)br(p)− dr(p)d†r(p)] (10.84)

If we now would assume commutation relations

[br(p), b†s(p

′)] = [dr(p), d†s(p

′)] = δrsδp,p′ (10.85)

the Hamiltonian could be rewritten, apart an infinite term, as

H =∑rp

E[b†r(p)br(p)− d†r(p)dr(p)] (10.86)

This Hamiltonian is unbounded from below and therefore the theory does notadmit a stable minimum. However, if we assume anticommutation relations

[br(p), b†s(p

′)]+ = [dr(p), d†s(p

′)]+ = δrsδp,p′ (10.87)

and all other anticommutators vanishing, the expression for H is

H =∑rp

E[b†r(p)br(p) + d†r(p)dr(p)] (10.88)

which is now definite positive and admits a minimum state |0 > with zeroenergy. The state |0 > is defined by the conditions

br(p)|0 >= dr(p)|0 >= 0 (10.89)

66

Dirac quantized theory describes two types of particles: one can build oneparticle states as

b†r(p)|0 >, and d†r(p)|0 > (10.90)

To distinguish these two types of particles we can consider additionaloperators commuting with H. Since the theory is invariant under gaugetransformations

ψ → eiαψ, ψ† → e−iαψ† (10.91)

we can build the corresponding Noether current (see eq.(3.79))

jµ =∂L∂∂µψ

∆ψ +∆ψ∂L∂∂µψ

(10.92)

which turns out to bejµ = ψ(x)γµψ(x) (10.93)

This current can also be identified by introducing in the Dirac Lagrangianthe interaction with the electromagnetic field by means of the minimal sub-sitution i∂µ → i∂µ − eAµ as done in Section 10.4.

Let us compute the total charge as

Q =

∫d3xj0 =

∫d3xψ†(x)ψ(x) (10.94)

By substituting the expansion (10.55) in Q we obtain, again neglecting aninfinite term,

Q =∑rp

[b†r(p)br(p)− d†r(p)dr(p)] (10.95)

Therefore the two types of particles have opposite charges. In conclusion theDirac equation describes the electron and its antiparticle the positron. Thisparticle was discovered in 1932 by the american physicist Carl Anderson whoreceived the Nobel prize in Physics in 1936. This discovery had been madeearlier by P. Blackett9 and G. Occhialini10 who however did not immediatelypublish their results. Dirac was awarded of the Nobel in 1933 together withSchrodinger “for the discovery of new productive forms of atomic theory”.

9Patrick Blackett, 1897-197410Giuseppe Occhialini,1907-1993

67

Quantization of a field theory with anticommutators implies Fermi Paulistatistics: the two particle state is antisymmetric under the exchange of thetwo particles

b†p1,rb†p2,s|0 >= −b†p2,sb

†p1,r|0 > (10.96)

Furthermore since [b†p,r]2 = 0 it is impossible to build a state with two elec-

trons with the same quantum numbers.

Using the invariance of the action under translation and Lorentz trans-formations on can also consider the total spatial momentum

P = −i∫d3xψ†∇ψ =

∑rp

p[b†r(p)br(p) + d†r(p)dr(p)] (10.97)

and the angular momentum:

M =

∫d3xψ†(x)[x× (−i∇)]ψ(x) +

∫d3xψ†(x)

1

2σψ(x) (10.98)

where σ a 4 by 4 matrix given by

σ =

(σ 00 σ

)(10.99)

The two terms represent the angular momentum and the spin term for 1/2particles.

Finally let us note that using the anticommutation relations (10.87) andthe expansion (10.55), one can derive the canonical anticommutation rela-tions

[ψ(t,x),Π(t,y)]+ = iδ3(x− y), [ψ(t,x), ψ(t,y)]+ = [Π(t,x),Π(t,y)]+ = 0(10.100)

11 Higgs decay width to fermions

As a simple application of perturbative approach of Quantum Field Theorywe consider the decay rate of the Higgs11 in electron and positron, H → e+e−.

11P. Higgs (1929-), Nobel prize in Physics in 2013

68

The Higgs is a spin zero particle, present in the Standard Model (SM) ofelectroweak interactions and responsible for the mechanism of generating themasses of the quarks u, d, c, s, t, b, of the leptons e, µ, τ, νe, νµ, ντ and of thegauge fields W±, Z. Quarks and leptons have spin 1/2 and thetefore can bedescribed by Dirac fields. The Higgs is not a stable particle and decays invarious channels. His main decay is in the quark channel bb where b denotesthe antiparticle of the quark b. The branching ratio in the bb channel isdefined as

B(H → bb) =Γ(H → bb)

ΓtotH(11.1)

where ΓtotH is the total width of the Higgs, which, within the SM, is predictedto be ΓtotH = 4.15 MeV. The branching ratio in the bb channel is approxi-mately 0.89 ± 0.29.

However this channel has a large background from strong interactions.The relevant decay channels for Higgs discovery are

γγ,W+W−, ZZ, τ+τ−, bb (11.2)

In the following we compute the H → e+e− width and then with a suitablerescaling the H → bb width.

The Higgs interaction with the spin 1/2 particles (quarks and leptons) isdecribed by an Hamiltonian

HI =

∫d3xHI (11.3)

withHI = −λφ(x)ψ(x)ψ(x) (11.4)

where φ(x) is the scalar field which describes the Higgs and ψ(x) the Diracfield describing the fermion. The interaction coupling is

λ =m

v∼ m

250 GeV(11.5)

where m is the mass of the field ψ. The coupling constant λ is dimensionless.In fact, assuming the mass as fundamental dimension, the dimensions of thescalar field are M1, the dimensions of the fermion field M3/2 and since

HI = −LI (11.6)

69

the dimensions of the lagrangian are M4 so that the action is dimensionless.In eq. (11.5) the parameter v ∼ 250 GeV is related to the energy scale (the

inverse of the range) of the weak interactions, G−1/2F . The Fermi constant

GF is related to v by

GF =1√2v2

(11.7)

The masses of the three gauge bosons W±, Z which mediate the weak inter-actions are of the order of v.

For the electron, me ∼ 0.5 MeV, λ ∼ 10−6, therefore the decay H → e+e−

is very small. However in the case of the µ, mµ = 105.6 MeV ∼200 me andthe decay rate is larger; LHC has already some candidates for the decayH → µ+µ−. For the calculation of the decay widths we will consider thefields quantized in a box of volume V . Therefore we have the followingexpansions

φ(x) =∑k

1√2EkV

(ake−ikx + h.c.)

ψ(x) =∑kr

√m

EkV[br(k)ur(k)e

−ikx + d†r(k)vr(k)eikx]

(11.8)

where we recall thatkx ≡ Ekt− k · x (11.9)

and Ek =√k2 +m2 where m denotes the corresponding mass of the boson

or of the fermion.

The initial state is therefore a Higgs with momentum p

|i〉 = a†p|0〉 (11.10)

while the final state is a positron with momentum and k1 and an electronwith momentum k2

|f〉 = d†r1(k1)b†r2(k2)|0〉 (11.11)

where r1, r2 are the spin labels. The vacuum is the direct product of thevacua of the two Fock (1934) spaces

|0〉 ≡ |0〉φ ⊗ |0〉ψ (11.12)

70

We have, evaluating the Hamiltonian density at t = 0,

〈f |HI |i〉 = −λ√

me

Ek1V

√me

Ek2V

1√2EpV

ei(p−k1−k2)·xur2(k2)vr1(k1) (11.13)

and

Vfi =

∫d3x〈f |HI |i〉 = −λ

√me

Ek1V

√me

Ek2V

1√2EpV

ur2(k2)vr1(k1)V δp,k1+k2

(11.14)

Using the golden rule, we can compute the probability that the Higgsof momentum p decays in a positron of momentum k1 and an electron ofmomentum k2

dwfi = 2πδ(Ek1 + Ek2 − Ep)|Vfi|2Vd3k1(2π)3

Vd3k2(2π)3

= (2π)4δ4(Pf − Pi)V λ2(

√me

Ek1V

√me

Ek2V

1√2EpV

)2

|ur2(k2)vr1(k1)|2Vd3k1(2π)3

Vd3k2(2π)3

(11.15)

where passing to the continuum we have used

(V δp,k1+k2)2 → (2π)3δ3(p− k1 − k2)V (11.16)

Pf,i are the final and initial total four-momentum. In order to compute thetotal width Γ(H → e+e−), we can now integrate over k1 and k2∫

dwfi =1

(2π)2λ2

2

∫d3k1d

3k2δ4(k1 + k2 − p)

m2e

Ek1Ek2Ep

|ur2(k2)vr1(k1)|2

=1

(2π)2λ2

2

∫d3k1δ(Ek1 + Ek2 − Ep)

m2e

Ek1Ek2Ep

|ur2(k2)vr1(k1)|2 (11.17)

and sum over r1 and r2.

71

Γ(H → e+e−) =∑r1r2

∫k1,k2

dwfi

=1

(2π)2λ2

2

∫d3k1δ(Ek1 + Ek2 − Ep)

m2e

Ek1Ek2Ep∑r1r2

vr1(k1)ur2(k2)ur2(k2)vr1(k1)

=1

(2π)2λ2

2

∫d3k1δ(Ek1 + Ek2 − Ep)

m2e

Ek1Ek2Ep1

4m2e

Tr[(k1 −me)(k2 +me)]

=λ2

8π2

∫d3k1δ(Ek1 + Ek2 − Ep)

k1 · k2 −m2e

Ek1Ek2Ep(11.18)

In the previous equations we have used∑r1r2

vr1(k1)ur2(k2)ur2(k2)vr1(k1) =∑r1

vr1β(k1)vr1α(k1)∑r2

ur2α(k2)ur2β(k2)

= −Tr(Λ−Λ+) (11.19)

where Λ± = 12me

(±k +me) are the positive (negative) energy solution pro-jectors. We have also used

Tr(γµγν) = 4gµν , Tr(γµ) = 0, Tr(I) = 4 (11.20)

Usually one computes the decay rate in the rest frame of the decayingparticle. Therefore

p = (mH ,0), k1 = (Ek1 ,k1), k2 = (Ek2 ,−k1) (11.21)

with

Ek1 = Ek2 =√m2e + k2

1 (11.22)

We havek1 · k2 = 2k2

1 +m2e (11.23)

72

Substituting in the width, after integration over dΩ we get

Γ(H → e+e−) =λ2

mHπ

∫ ∞

0

dk1k41

k21 +m2e

δ(mH − 2√k21 +m2

e)

=λ2

mHπ

∫ ∞

0

dk1k31√

k21 +m2e

δ(k1 −1

2

√m2H − 4m2

e)

=λ2

8πmH(1−

4m2e

m2H

)3/2 (11.24)

where in the last equation we have used the property

δ(f(x)) =1

|f ′(x0)|δ(x− x0) (11.25)

where x0 is a zero of f(x) of order one. Using mH ∼ 125 GeV and me ∼0.5 MeV we get

Γ(H → e+e−) ∼ 2× 10−11 GeV = 2× 10−2 eV (11.26)

For the quark bottom (mb = 4.2 GeV) we obtain, taking into account afactor three from the sum over the three colors of the bottom quark (thecolor gauge interaction which is responsible of the strong force is based onthe SU(3) group and the b quark is a triplet of SU(3))

Γ(H → bb) ∼ 3(mb

me

)2Γ(H → e+e−) ∼ 4 MeV (11.27)

This is the main decay channel of the Higgs. However this channel is notthe cleanest because of QCD background. The main discovery channels atLHC were γγ, ZZ∗ and W+W−∗ where the ∗ denotes a virtual particle. Forexample, when considering H → ZZ since mH ≤ 2mZ only one Z particlecan be real. The second Z is only a virtual intermediate state decaying intwo particles which are detected in the detectors ATLAS and CMS of LHC.

12 The decay width for the process µ− →e−νeνµ

The decay µ− → e−νeνµ is due to the weak interactions and can be explainedby the Fermi interaction12 between the Dirac fields representing the muon,

12The theory was proposed by Fermi in 1934

73

the electron and the two neutrinos.The effective Hamiltonian density is givenby a current current interaction

HW =GF√2(J†λe Jµλ + h.c) (12.1)

where the electronic and the muonic currents are

Jλe = ψeγλ(1− γ5)ψνe , J

λµ = ψµγ

λ(1− γ5)ψνµ (12.2)

The Fermi constant GF can be derived, as we will see, by the experimentalvalue of the decay width and turns out to be GF = 1.16 × 10−5 GeV−2.Remember that each fermion field ψ describes both particle and antiparticles

ψ(x) = ψ(+)(x) + ψ(−)(x) (12.3)

where ψ(±) denote the positive and negative energy part of the spinor.

The initial state is a muon with given momentum and spin

|µ−(pµ, rµ) >≡ |µ−(pµ, rµ) > ⊗|0 >e− ⊗|0 >νµ ⊗|0 >νe (12.4)

and the final state is

|e νµνe >= |0 >µ ⊗|e−(pe, re) > ⊗|νµ(pνµ , rνµ) > ⊗|νe(pνe , rνe) > (12.5)

Each fermion field can be expanded as

ψ(x) =1√V

∑r

∑p

√m

E[br(p)ur(p)e

−ipx + d†r(p)vr(p)eipx] (12.6)

One can verify that

ψ(+)µ (x)|µ−(pµ, rµ) >=

1√V

√mµ

Eµurµ(pµ)e

−ipµx|0 > (12.7)

< e(pe, re)|(ψ(+)e )†(x) =

1√V

√me

Eeu†re(pe)e

ipex < 0| (12.8)

and similarly for νe and νµ.

< νe(pνe , rνe)|(ψ(−)νe )(x) =

1√V

√mνe

Eνevrνe (pνe)e

ipνex < 0| (12.9)

74

< νµ(pνµ , rνµ)|(ψ(+)νµ )†(x) =

1√V

√mνµ

Eνµu†rνµ (pνµ)e

ipνµx < 0| (12.10)

where|0 >≡ |0 >µ ⊗|0 >e− ⊗|0 >νµ ⊗|0 >νe (12.11)

The relevant term of the interaction (12.1) is

J†µλJ

λe = ψ†

νµ(1− γ5)γ†λγ0ψµψ

†eγ0γ

λ(1− γ5)ψνe

= ψ(+)†νµ (1− γ5)γ

†λγ0ψ

(+)µ ψ(+)†

e γ0γλ(1− γ5)ψ

(−)νe + . . . (12.12)

Therefore we obtain, by considering the interaction Hamiltonian density att = 0,

< f |HW |i > =GF√2< e νµνe|J†

µλJλe |µ >

=GF√2

√mµ

Eµ

√me

Ee

√mνe

Eνe

√mνµ

Eνµ

1

V 2ei(pνµ+pνe+pe−pµ)x|x0=0

u†re(pe)γ0γλ(1− γ5)vrνe (pνe)u†rνµ

(pνµ)γ0γλ(1− γ5)urµ(pµ)

=GF√2

√mµ

Eµ

√me

Ee

√mνe

Eνe

√mνµ

Eνµ

1

V 2ei(pνµ+pνe+pe−pµ)x|x0=0

urne(pe)γλ(1− γ5)vrνe (pνe)urνµ (pνµ)γ

λ(1− γ5)urµ(pµ)

and finally

Vfi =

∫d3x < f |HW |i >

=GF√2

√mµ

µ

√me

Ee

√mνe

Eνe

√mνµ

Eνµ

1

V 2V δ3(pνµ + pνe + pe − pµ)

urne(pe)γλ(1− γ5)vrνe (pνe)urνµ (pνµ)γ

λ(1− γ5)urµ(pµ)

≡ MfiGF√2

√mµ

Eµ

√me

Ee

√mνe

Eνe

√mνµ

Eνµ

1

Vδ3(pνµ + pνe + pe − pµ)

(12.13)

where

Mfi = urνe (pe)γλ(1− γ5)vrνe (pνe)urνµ (pνµ)γλ(1− γ5)urµ(pµ) (12.14)

75

Let us now compute the decay rate, using the Fermi golden rule and averagingover the initial spin and summing over the final spin

dw =1

2

∑ri,rf

2πδ(Ef − Ei)|Mfi|2|1

V

√mµ

Eµ

√me

Ee

√mνe

Eνe

√mνµ

Eνµ|2

(2π)3

Vδ3(pνµ + pνe + pe − pµ)

V

(2π)3d3pe

V

(2π)3d3pνe

V

(2π)3d3pνµ

(12.15)

Therefore the final result can be written as

dw = (2π)4δ4(Pf − Pi)mµ

EµΠf

d3pf(2π)3

mf

Ef

1

2

∑ri,rf

|Mfi|2 (12.16)

with (see Appendix F)

1

2

∑ri,rf

|Mfi|2 = 64G2F

1

2mνµ2mνe2mµ2me

(pµ · pνe)(pνµ · pe) (12.17)

This can be rewritten as

dw = (2π)4δ4(Pf − Pi)Πfd3pf

2Ef (2π)31

2Eµ< M >2 (12.18)

with< M >2= 64G2

F (pµ · pνe)(pνµ · pe) (12.19)

The total width is computed in the rest frame of the muon

pµ = (mµ,0) (12.20)

In this frame we have(pµ · pνe) = mµEνe (12.21)

Furthermore from

(pµ − pνe)2 = (pνµ + pe)

2 (12.22)

76

neglecting neutrino and electron masses we get

(pνµ · pe) ∼1

2m2µ −mµEνe (12.23)

and therefore

< M >2∼ 64G2F

m2µ

2Eνe(mµ − 2Eνe) (12.24)

The first integration over pνµ is trivial and we obtain∫pνµ

dΓ =1

16(2π)5mµ

d3pνeEνe

d3peEeEνµ

δ(mµ − Eνe − Eνµ − Ee) < M >2 (12.25)

with < M >2 given in (12.24). In the µ rest frame we have pµ = 0, pνµ =−pνe − pe. Then from

Eνµ = |pνµ| = |pνe + pe| (12.26)

we get

Eνµ =√E2νe + E2

e + 2EνeEe cos θ (12.27)

Also we haved3pνe = E2

νedEνe sin θdθdφ (12.28)

To perform the integral over θ we can transform from θ to

u ≡ Eνµ =√E2νe + E2

e + 2EνeEe cos θ (12.29)

So

du = −EνeEe sin θdθEνµ

(12.30)

and we get∫pνe ,pνµ

dΓ =

∫dEνe < M >2 d3pe

1

16(2π)4mµE2e

∫ u(cos θ=1)

u(cos θ=−1)

duδ(mµ−Eνe−u−Ee)

(12.31)Let us discuss the extrema of the integration

u(cos θ = 1) = |Eνe + Ee| (12.32)

77

u(cos θ = −1) = |Eνe − Ee| (12.33)

So the integral is different from zero when

|Eνe − Ee| < u = mµ − Eνe − Ee < |Eνe + Ee| (12.34)

which are equivalent to

|Eνe − Ee|+ Eνe + Ee <mµ

2< Eνe + Ee (12.35)

This implies

Eνe <mµ

2< Eνe + Ee (12.36)

when Eνe > Ee and

Ee <mµ

2< Eνe + Ee (12.37)

when Eνe < Ee. Therefore the extrema are

mµ

2− Ee < Eνe <

mµ

2(12.38)

By integrating over dEνe∫dΓ =

∫ mµ2

mµ2

−Ee

dEνe < M >2 d3pe1

16(2π)4mµE2e

= 2G2F

d3pe(2π)4

(mµ

2− 2Ee

3) (12.39)

Then we can integrate over pe.

Γ =2G2

F

(2π)4mµ

∫ mµ/2

0

dEeE2e

∫Ω

sin θdθ(mµ

2− 2Ee

3)dφ (12.40)

The final result is

Γ(µ→ e−νeνµ) =G2Fm

5µ

3× 26π3(12.41)

The decay time is given by

τµ =3× 26π3

m5µG

2F

(12.42)

78

Using the experimental determination13 τµ ∼ 2.2× 10−6sec and mµ ∼ 105.7MeV, one can extract the value of the Fermi constant

GF = 1.16× 10−5GeV−2 (12.43)

13PDG, Particle Data Group: τµ = (2.1969811 ± 0.0000022 × 10−6) sec, mµ =(105.6583745 ± 0.0000024) MeV . The first measurements of the µ life time were per-formed by F. Rasetti (1941), τµ = (1.5 ± 0.3 × 10−6) sec, and B. Rossi (1943), τµ =(2.15± 0.07× 10−6) sec

79

13 Superfluidity

In my recent studies on liquid heliumclose to the absolute zero, I have succeeded in discovering a num-ber of new phenomena. . . I am planning to publish part of thiswork. . . but to do this I need theoretical help. In the SovietUnion it is Landau who has the most perfect command of the the-oretical field I need, but unfortunately he has been in custody fora whole year. All this time I have been hoping that he would bereleased because, frankly speaking, I am unable to believe that heis a state criminal. . .” P. Kapitza, letter to Molotov on April6,1939

80

13.1 Brief introduction to superfluids

There are two stable isotopes of the helium: the first He4 was discovered in1871 in the solar spectrum, while the He3 was discovered in 1939 at LBL byLouis Alvarez14.

The isotope He4, which has a nucleus composed of two protons and twoneutrons, is a boson (spin 0) while He3 composed of two protons and oneneutron, is a fermion (spin 1/2). Both liquids have, at low temperatures, lowdensities and apparently remain liquid down to absolute zero temperatureunless a high pressure is applied (25 atm for the He4). The density of theHe4 is

ρ4 = 0.145g/cm3 (13.1)

at T = 2.17 K and p = 0.0497 atm.

This property of not freezing comes from the extremely weak Van derWalls forces between the atoms with respect to the quantum fluctuations.

The two liquids behave in different way because the Pauli principle keepsHe3 fermions far each other. The He4, below Tλ =2.17 K, enters in a newphase, HeII, Fig. 1.

The first researcher who liquified the Helium below the gas liquid tran-sition at 4.2 K was Kamerlingh Onnes15 (1908), in the experiments leadingto discover superconductivity. Later in 1927, M. Wolfke and W.H. Keesomdiscovered a new phase transition at 2.17 K that manifested itself in a dis-continuity of the specific heat. After this observation, it took 10 years tounderstand that the new phase was a superfluid phase: the fluid can flowwithout any friction and viscosity.

The transition line λ is the separation between HeI and HeII phases, thefirst a normal liquid, the second superfluid. In this phase the liquid is capableof flowing through narrow capillaries without friction. Experiments to provesuperfluidity where first performed by Kapitza16 (1938) and indipendentlyby J.F. Allen and A.D. Misener (1938).

14Nobel prize in Physics in 1968 for his contribution to particle physics, in particularthe bubble chamber

15H. Kamerlingh Onnes, 1853-1926, Nobel prize in Physics in 191316P.L. Kapitza, 1894-1984, Nobel prize in Physics in 1978

81

Figure 1: Phase diagram of superfluid He-4 (from J.C.Davis Group, Cornell)

Figure 2: Specific heat (from Huang)

82

It is natural to associate the λ transition to the Bose Einstein condensa-tion (London 1938) modified by the molecular interactions. In fact, as wewill review in the following, the critical temperature for the condensation ina non interacting gas of bosons is 3.14 K, very close to Tλ.

The He3 also becomes superfluid but only at temperatures of the or-der 10−3K (1972, Lee, Osheroff and Richardson, 17): in this case pairs offermions condensate, a similar mechanism to the Cooper pairs for supercon-ductors. Recently, in 2005, the superfluidity has been observed in ultracoldFermi gases (Ketterle18 et al) at very low temperature, 200 nK.

H2 solidifies at higher temperature because of stronger interactions amonghis molecules.

An additional property of the Helium is that at low temperature (T <<Tc) the specific heat varies as T3, as shown in Fig. 2.

To explain such a behavior Landau19 (1941) suggested to interpret thequantum states of the liquid as a phonon gas with the linear dispersionrelation

εk = ~ck (13.2)

where c is the velocity of propagation of the sound in the fluid. The mainidea is that the body moving in the liquid excites sound waves which arecollective motions in the liquid. The liquid at low temperature must betreated as a quantum liquid and its excitations are phonons as for the quan-tum excitations of the vibrations of a crystal. A body moving in the heliumat low temperature does not loose energy transferring to single atoms butexcites collective quanta as phonons. The helium dispersion relation curve ismeasured by neutron scattering, see Fig. 3. Approximately one has

εk = ~ck, k << k0 (13.3)

with c = (239± 5) m/s

εk = ∆+~2(k − k0)

2

2σ, k ∼ k0 (13.4)

17Nobel prize in Physics in 197818W. Ketterle, 1957- Nobel prize in Physics in 200119L. Landau, 1908-1968, Nobel prize in Physics in 1962

83

and with ∆/kB = (8.65 ± 0.04) K, with kB the Boltzmann constant, k0 =(1.92 ± 0.01)108 cm−1, σ/m = 0.16 ± 0.01 with m the mass of the heliumatom. Therefore the dispersion relation is linear for small k and has a localminimum at k = k0.

The nature of the excitations in the helium is studied by measuring thechange in energy and momentum in cold neutron scattering on Helium su-perfluid (see for example, Palewsky et al, Physical Review 112, (1958), 11).The reason is that cold neutrons have momentum close to the momentum ofthe excitations (the energy of the neutron is ∼ 50K).

Making use of the conservation laws

1

2m(p2

i − p2f ) = ε(k) (13.5)

pi − pf = ~k (13.6)

where pi(f) are the momenta of the incoming (outgoing) neutron one canobtain the spectrum of the excitations.

Let us now see how it is possible that an object can move in a superfluidwithout loosing energy (Landau criterion for superfluidity). Let us consideran object of mass M moving in a superfluid. The only way in which it canloose energy is by emission of a phonon

P2

2M− (P− ~k)2

2M= −~2k2

2M+

P · k~M

= εk (13.7)

Therefore

(V · k)~ = εk +~2k2

2M≥ c~k (13.8)

or~V k ≥ |V · k| ≥ ck (13.9)

implyingV ≥ c (13.10)

Therefore if V ≤ c the process is prohibited. This explanation depends inan essential way from the linear spectrum of the phonons. If the spectrumof the excitations is quadratic the minimum threshold for loosing energy iszero.

84

At energies around k0 the object looses energy by emitting the so-calledrotons:

(V · k0)~ = εk0 +~2k2

2M≥ εk0 (13.11)

orV k0 ≥ |V k0~ cos θ| ≥ εk0 = ∆ (13.12)

orV ≥ vc (13.13)

with vc = ∆/(~k0) = 8.65 1.38 10−23J/(1.055 10−34Js1.92) × 10−8 cm∼ 58m/s. The rotons are the elementary excitations associated to vortices inthe fluid. At low temperature the roton effects are negligible due to theBoltzmann factor exp (−∆/KT ). At the thermal equilibrium elementaryexcitations have energies close to the minimum of ε that is zero. In presenceof a purely phonon spectrum the critical velocity is c = 239 m/sec, whenrotons are included the critical velocity drops to vc = 58 m/sec (observed inHe4 under pressure).

For a general spectrum εk the condition is

V ≥ εk~k

(13.14)

Therefore the critical velocity, defined as

Vc = min(εk~k

) (13.15)

for a free particle spectrum, ε(p) = ~2k2/2m is zero.

13.2 Bose Einstein condensation for an ideal gas

Since the He4 atom contains six spin 1/2 particles which are bound in sucha way that the total spin is zero, we can consider an ideal gas consisting ofN bosons in a volume V . The partition function is given by

ΩB = kBT∑p

log [1− exp β(εp − µ)] (13.16)

85

Figure 3: Experimental data and fit to the neutron scattering data, FromR.J. Donnelly, J.A. Donnelly and R.N. Hills, Journal of Low temperaturePhysics, 44 1981 471

86

where µ is the chemical potential. Let us now recall the average number of anideal gas of N non interacting bosons, derived by using the grand partitionfunction,

NB = −∂Ω∂µ

=∑p

1

exp β(εp − µ)− 1(13.17)

and passing to the continuum

N = N0 +V

h3

∫d3p

1

exp [β(εp − µ)]− 1(13.18)

where N0 is the number of bosons with p = 0.

At T = 0 all bosons occupy the state with p = 0 (Bose condensation).At finite temperature, T 6= 0, only a fraction N0/N of bosons remain in thestate p = 0. For T >> Tc, where Tc is the critical temperature, there is nocondensate, N0 = 0, n = N/V requiring µ < 0 because of the singularity forµ > 0. When T decreases, for fixed N/V , the absolute value of the chemicalpotential decreases until for temperatures sufficiently low reaches the valueof 0. The condensation starts when µ = 0

N0

N= 0 µ = 0 (13.19)

or

N = 0 +V

h3

∫d3p

1

exp[ εpkTc

]− 1

=V

h3(2mkTc)

3/24π

∫ ∞

0

x21

ex2 − 1dx

=V

λ3cg3/2(1) (13.20)

where

λc =

√2π~2mkBTc

(13.21)

From this condition we can get the critical temperature

Tc =1

kB

2π~2/m[vg3/2(1)]2/3

=1

kB

2π~2/m[ζ(3/2)]2/3

(N

V)2/3 (13.22)

87

where ζ is the Riemann function (see Appendix G). Using the numer-ical values ρHe4 = 0.145g/cm3 = mHe4N/V with mHe4 = 4mp, mp =4 × 1.67 × 10−27Kg, one can get N/V . Then inserting ~ = 1.05510−34 Jsec, the Boltzmann constant kB = 1.38 10−23J/K, ζ(3/2) = 2.61, we getTc = 3.14 K. This temperature is very close to the critical temperature ofliquid Helium, Tλ = 2.17 K, below which the helium becomes superfluid.

For T < Tc µ remains zero and using eq. (13.18), with µ = 0 we get

N −N0

V=

∫d3p

h31

exp [εp/kT ]− 1

=1

λ3g3/2(1) =

(mkT

2π~2

)3/2

g3/2(1) =N

V

(T

Tc

)3/2

(13.23)

orN −N0

N=

(T

Tc

)3/2

(13.24)

orN0

N= 1−

(T

Tc

)3/2

(13.25)

andN0

V=N

V

[1−

(T

Tc

)3/2]T < Tc (13.26)

In conclusion below Tc a fraction of particles occupy the state with p = 0.Therefore for T < Tc we have a condensate with a macroscopic number ofparticles in the same quantum state with p = 0. Bose Einstein condensationprovides only a qualitative description of superfluidity. For example thespecific heat of a non interacting boson gas vanishes as T 3/2 while the specificheat of He4 behaves as T 3. Furthermore we have superfluidity only for zerovelocity of the atoms given that the spectrum is the free particle one, ε =p2/2m.

13.3 Quantization of the Schrodinger field

Let us now see how to reformulate the non relativistic Schrodinger theory inthe second quantization formalism, or in the quantum field theory language.

88

Let us consider a non relativistic particle with zero spin which satisfies theSchrodinger equation:

i~∂

∂tψ = Hψ (13.27)

con H = p2/2m. Quantum states can be built by working in the Fock space:by postulating the existence of creation and destruction operators a†k e akwhich satisfy the commutation relations

[ak, ak′ ] = [a†k, a†k′ ] = 0 [ak, a

†k′ ] = δk,k′ (13.28)

In general, the one particle state is given by

|ψ >=∑k

ψka†k|0 > (13.29)

In this basis the wave function is given by

ψ(x) =< x|ψ >=∑k

ψk < x|a†k|0 >==∑k

ψk1√Veik·x (13.30)

We can now consider the following field

Φ(x, 0) =1√V

∑k

eik·xak (13.31)

such that

|ψ >=∫d3xψ(x)Φ†(x, 0)|0 > (13.32)

We have∫d3xψ(x)Φ†(x, 0)|0 >=

∫d3x

1

V

∑k′

ψk′eik′·x

∑k

e−ik·xak|0 >=∑k

ψka†k|0 >

(13.33)In general

Φ(x, t) =1√V

∑k

ei(k·x−ωkt)ak (13.34)

with ωk = ~k2/2m. The field Φ(x, t) satisfies the Schrodinger equation

i~∂

∂tΦ = −~2

∇2

2mΦ (13.35)

89

Let us now write the Lagrangian associated to the Schrodinger equation

L = i~Φ†Φ− ~21

2m∇Φ† · ∇Φ (13.36)

and the corresponding Hamiltonian density

H = ΠΦ− L = ~21

2m∇Φ† · ∇Φ (13.37)

with Π = ∂L/∂Φ = i~Φ†. The commutation relations (13.28) imply

[Φ(x, t),Π(y, t)] = i~δ3(x− y) (13.38)

The Hamiltonian of the field is given by

H =

∫d3xH =

∑k

~ωka†kak (13.39)

As an application of the non relativistic field theory we will consider thesuperfluidity theory.

Bogoliubov (1947) studied the fundamental state of a dilute gas of weaklyinteracting bosons and their excitations using the second quantization of amany body system and assuming the following interaction Hamiltonian, see[9, 10]:

HI =1

2

∑k1+k2=k′

1+k′2

W (|k1 − k′1|)a

†k′1a†k′

2ak1ak2 (13.40)

where ~k1, ~k2 (~k′1, ~k′

2) represent the momenta of incoming (outgoing)bosons and ~k′

1 − ~k1 the transfer momentum. The function W (k) is theFourier transform of the four boson interaction

W (k) =

∫d~rW (r)eik·r (13.41)

The dilute gas approximation justifies to consider only two boson scattering.

Before discussing the Bogoliubov approach let us first review how one candescribe the superfluidity phase by use of the general approach of Ginzburg20-Landau to phase transitions.

20V.L. Ginzburg, 1916-2009, Nobel prize in Physics in 2003

90

13.4 Ginzburg-Landau Model

The superfluidity phase transition can be derived within the Ginzburg-Landauapproach to phase transitions, assuming that the states of the system are de-scribed by a scalar field which can be interpreted as the wave function of thesuperfluid.

The Hamiltonian of the model (the free energy), which was proposed asan effective description of field theory for phase transitions, is given by

Heff =~2

2m(∇φ)†(∇φ)− µφ†φ+

1

2g(φ†φ)2 (13.42)

with µ and g positive constants. In particular µ < 0 for T > Tc and passesthrough zero at Tc.

Therefore the potential can be identified as

V (φ) = −µφ†φ+1

2g(φ†φ)2 (13.43)

and is dominated for low density by the chemical potential and at largedensity by the g term. The form of the potential is shown in Fig. 4. Amicroscopic interpretation of the parameters µ and g can be found in [10].They can be related to the strength of the four boson interaction and to thedensity of the condensate.

Let us now assume that the chemical potential depends on the tempera-ture, so that µ < 0 for T > Tc, Tc being the critical temperature, and µ > 0for T < Tc. The potential, for T < Tc, has a maximum in |φ| = 0 and aminimum for

φ†φ =µ

g(13.44)

or

φ(x) = φ0 exp (iψ) =

õ

gexp (iψ), (13.45)

The minimum is degenerate varying ψ ∈ [0, 2π). For simplicity let us choosethe minimum at ψ = 0. The series of the field in normal modes can beperformed with respect to the new minimum in φ0

φ(x) = φ0 + φ(x) = φ0 +∑k 6=0

1√Vake

ik·x (13.46)

91

0.5 1.0 1.5 2.0ÈΦÈ

-0.2

-0.1

0.1

0.2

0.3

V

Figure 4: Potential V corresponding to eq.(13.43) for µ = 0.4 and g = 0.3 as

a function of |φ| =√φ†φ.

Let us notice that the Hamiltonian (13.42) is invariant under the transfor-mation

φ(x) → φ(x) exp (iα), α ∈ [0, 2π) (13.47)

while the minimum state is not (φ0 → φ0 exp (iα)). This phenomenon iscalled Spontaneous Symmetry Breaking an it is used for describing phasetransitions.

By substituting eq.(13.46) in the potential (13.43) one gets, by expandingto second order in φ

V = −µ[φ20 + φ0(φ+ φ†) + φ†φ] +

1

2g[φ40 + φ2

0(φ+ φ†)2 + 2φ30(φ+ φ†) + 2φ2

0φ†φ]

+O(φ3, φ4)

= −gφ20

[φ20 + φ0(φ+ φ†) + φ†φ

]+

1

2gφ4

0 +1

2gφ2

0(φ+ φ†)2 + gφ30(φ+ φ†)

+gφ20φ

†φ+O(φ3, φ4)

= −1

2gφ4

0 +1

2gφ2

0(φ+ φ†)2 +O(φ3, φ4)

= −µ2

2g+

1

2µ(φ+ φ†)2 +O(φ3, φ4) (13.48)

Let us now quantize the scalar field φ, by requiring standard commutationrelations ak e a†k. By substituting the normal mode series and integrating in

92

d3x the Hamiltonian density, one obtains

Heff =∑k6=0

[(µ+

~2k2

2m

)a†kak +

µ

2(aka−k + a†ka

†−k)

]+ E0 (13.49)

where

E0 = −µ2

2gV +

µ

2

∑k6=0

1 (13.50)

The Hamiltonian (13.49) is not diagonal in the basis of occupation num-bers because of the bi-linear terms a e in a†. However it is possible to find atransformation (Bogoliubov transformation, see Appendix I) from ak(a

†k) to

the operators Ak(A†k), defined as

Ak = cosh(θk2)ak + sinh(

θk2)a†−k (13.51)

withtanh θk =

µ

µ+ ~2k2

2m

(13.52)

One has[Ak, A

†~k′] = δk,~k′ (13.53)

FurthermoreHeff = Enew

0 +∑k6=0

ε(k)A†kAk (13.54)

with

Enew0 = E0 −

∑k6=0

ε(k) sinh2(θk2) (13.55)

and

ε(k) =

√(µ+

~2k2

2m

)2

− µ2 =

õ

m~2k2 +

(~2k2

2m

)2

(13.56)

The fundamental state is defined as

Ak|φ0 >= 0 (13.57)

93

Starting from this new vacuum state one can build the new Fock space withthe operators A†

k. For example, the first excited state (or quasi-particle) isgiven by

A†k|φ0 > (13.58)

with energy

ε(k) =

õ

m~2k2 +

(~2k2

2m

)2

(13.59)

Therefore the spectrum is linear for small k while for large k behaves as k2.

In conclusion the Ginzburg-Landau approach is able to explain the spec-trum of the liquid helium at low k but does not reproduce the local minimumdue to the rotons.

The relation of the new vacuum |φ0 > with the Fock vacuum state |0 >is given in Appendix H.

13.5 Bogoliubov approach

Let us first rewrite the Bogoliubov interaction Hamiltonian

HI =1

2

∑k1+k2=k′

1+k′2

W (|k1 − k′1|)a

†k′1a†k′

2ak1ak2 (13.60)

For T << Tc all bosons tend to belong to the state k = 0, thereforeN0 ∼ N and

N −N0

N0

<< 1 (13.61)

Since N0 ∼ N >> 1 one can assume the operator a0 to be c-number a0 ∼a†0 ∼

√N0. So we consider the operators ak and a†k small with respect to

a0 and a†0, expanding the interaction Hamiltonian keeping only the termswhich are linear or bi-linear in N0. In other words Bogoliubov separates thecondensate in the expansion of the field:

φ ∼ 1√V(√N0 +

∑k6=0

ak exp [i(k · x)]) (13.62)

94

Let us first rewrite the interaction Hamiltonian by taking into accountthe momentum conservation

HI =1

2

∑k1,k2,k′

1

W (|k1 − k′1|)a

†k′1a†k1+k2−k′

1ak1ak2 (13.63)

Let us first list all the cases where the indices are zero. One has four zeroindices for k1 = k2 = k′

1 = 0. One has two zero indices when (we list alsothe transfer momentum)

k1 = k′1 = 0 k1 − k′

1 = 0

k1 = k2 = 0 k1 − k′1 = −k′

1

k2 = k′1 = 0 k1 − k′

1 = k1

k2 = k1 − k′1 = 0 k1 − k′

1 = 0

k1 = k2 − k′1 = 0 k1 − k′

1 = −k′1

k′1 = k1 + k2 = 0 k1 − k′

1 = k1 (13.64)

Therefore neglecting all terms of order O(a4k) we obtain

HI =1

2

[W (0)(a†0a0)

2

+ W (0)a†0a0∑k2

a†k2ak2 + a0a0

∑k′1

W (k′1)a†k′1a†−k′

1

+ a†0a0∑k1

W (k1)a†k1ak1 +W (0)a†0a0

∑k1

a†k1ak1

+ a†0a0∑k′1

W (k′1)a†k′1ak′

1+ a†0a

†0

∑k1

W (k1)a−k1ak1

]=

1

2

[W (0)N2

0 + 2N0

∑k 6=0

(W (0) +W (k))a†kak

+ N0

∑k 6=0

W (k)(a†ka†−k + aka−k)

](13.65)

Furthermore the number operator

N = a†0a0 +∑k6=0

a†kak (13.66)

95

so that neglecting order O(a4k)

N20 ∼ N2 − 2N

∑k6=0

a†kak (13.67)

By substituting eq.(13.67) in eq.(13.65) we get

HI =1

2[W (0)N2 + 2N

∑k6=0

W (k)a†kak

+ N∑k6=0

W (k)(a†ka†k + akak)

](13.68)

The total Hamiltonian, obtained adding to HI the kinetic term, can be di-agonalized as shown in Appendix I by the Bogoliubov transformation givenin eq.(I.3)

tanh θk =NW (k)

NW (k) + ~2k22m

(13.69)

The total Hamiltonian can be rewritten as∑k

ε(k)A†kAk (13.70)

with

ε(k) =

√(NW (k) +

~2k22m

)2 − (NW (k))2

=

√(~2k22m

)2 +NW (k)~2k2m

(13.71)

With a suitable choice of W (k) one is able to reproduce not only thephonon part of the spectrum but also the roton part.

Finally, by comparison with the Ginzburg-Landau Hamiltonian, we ob-tain the identification

µ ∼ NW (0), g = W (0)V (13.72)

and so if we neglect the interaction, we recover µ = 0, or the vanishing of theµ parameter below the critical temperature in the free boson gas approachto superfluidity.

96

14 Superconductivity

14.1 BCS Hamiltonian

Let us now consider, as a second application of non relativistic quantum fieldtheory, the phenomenon of superconductivity. Superconductivity is charac-terized by two main properties:

• In many metals, for example lead, tin,aluminium, cadmium, niobium...below a critical temperature Tc ∼ few K, resistivity drops to zero(the discovery was made working at low temperature with mercury byKamerlingh Onnes, 1911)

• Meissner effect: exclusion of magnetic fields from superconducting re-gions. The magnetic field decreases exponentially over distances of

order 500A (Meissner, Ochsenfeld 21, 1933)

The theoretical explanation is based on the formation of Cooper22 pairs:below the critical temperature the interaction between electrons close to theFermi surface and the phonons of the ion lattice can compensate for theCoulomb repulsion and provides the mechanism for the formation of Cooperpairs. Cooper showed (1956) that the Fermi sea of electrons is unstableagainst the formation of Cooper pairs.

One can show that the excitations of such a system have a spectrumwhich has a minimum corresponding to a finite energy gap and therefore anelectron moving in the metal cannot loose energy if its energy is below thegap. Therefore the current flows without resistivity.

In the following we will follow the Bardeen23, Cooper, Schrieffer24 ap-proach (1957). We consider a non relativistic spinor field

ψσ(x, t) =1√V

∑k,σ

ck,σuσ exp [−i(ωkt− k · x)] (14.1)

21F.W. Meissner 1882-1974, R. Ochsenfeld, 1901-199322L. Cooper, 1930-, Nobel prize in Physics in 197223John Bardeen, 1908-1991, Nobel prize in Physics in 1956 and in 197224J. R. Schrieffer, 1931-, Nobel prize in Physics in 1972

97

where uσ, σ = 1, 2 are the two orthogonal two dimensional spinors and theoperators ck,σ, c

†k′,σ′ satisfy the anticommutation relations

[ck,σ, c†k′,σ′ ]+ = δk,k′δσ,σ′ (14.2)

The Bardeen, Cooper and Schrieffer (1957) Hamiltonian is given by thegrand canonical Hamiltonian which includes a term −µN where µ ∼ µF =p2F/2m. In other words the chemical potential is approximated by its valueat the Fermi surface. We have

H =∑k,σ

ξkc†k,σck,σ −

1

V

∑k,k′

Wk,lc†k,↑c

†−k,↓c−k′,↓ck′,↑ (14.3)

with

ξk = εk − εF =~2k2

2m− ~2k2F

2m(14.4)

and Wkk′ =W 6= 0 only for electrons close to the Fermi surface

|ξk|, |ξk′| ≤ ~ωD (14.5)

where ωD is the Debye frequency and ~ωD can be considered as an estimateof the phonon energy, otherwise Wkk′ = 0. This can be understood from thefact that only electrons close to the Fermi states can scatter from a phononand find a different and not occupied final state. The shell is very tinyωD/ωF ∼ 10−3.

Since at low temperature the phonon electron interaction generates acondensate with pairs of electron of opposite spin and momentum the newvacuum (fundamental state) of the theory must be such that

< c−k,↓ck,↑ > 6= 0 and < c†−k,↓c†k,↑ > 6= 0 (14.6)

Therefore the new vacuum |BCS > 6= |0 > since the standard vacuum |0 >satisfies

ck,σ|0 >= 0 (14.7)

Let us now see whether it is possible to find a new vacuum |BCS > suchthat

< BCS| : c−k,↓ck,↑ : |BCS >= 0 (14.8)

98

and< BCS|c−k,↓ck,↑|BCS >= Qk 6= 0 (14.9)

orc−k,↑ck,↓ = Qk+ : c−k,↑ck,↓ : (14.10)

As we have done in superfluidity we perform the transformation from ck,↑, c−k,↓to a new pair of operators Ak, Bk

Ak = ukck,↑ − vkc†−k,↓

Bk = ukc−k,↓ + vkc†k,↑ (14.11)

where we assume uk, vk real.

By requiring the anticommutation relation for Ak, Bk

[Ak, A†k′ ]+ = [Bk, B

†k′ ]+ = δk,k′ (14.12)

we getu2k + v2k = 1 (14.13)

which can be satisfied assuming

uk = cos θk , vk = sin θk (14.14)

We require alsoAk|BCS >= Bk|BCS >= 0 (14.15)

By using the inverse relations

ck,↑ = ukAk + vkB†k

c−k,↓ = −vkA†k + ukBk (14.16)

we get

c−k,↓ck,↑ = (−vkA†k + ukBk)(ukAk + vkB

†k) = ukvk+ : c−k↓ck↑ : (14.17)

with

: c−k↓ck↑ : = −ukvk(A†kAk +B†

kBk)

+ u2kBkAk − v2kA†kB

†k (14.18)

99

We can now perform the transformation in the Hamiltonian: first the kineticterm∑

k

ξk(c†k↑ck↑ + c†k↓ck↓) = 2

∑k

ξkv2k +

∑k

ξk(u2k − v2k)(A

†kAk +B†

kBk)

− 2∑k

ξkukvk(B†kA

†k + AkBk) (14.19)

and then the interaction term, neglecting terms of order O(c4k),

− 1

V

∑k,k′

Wkk′c†k,↑c†−k,↓c−k′,↓ck′,↑ = − 1

V

∑k,k′

Wkk′ukvkuk′vk′

− 1

V

∑k

ukvk∑k′

Wkk′(: c†k′,↑c†−k′,↓ : + : c−k′,↓ck′,↑ :)

(14.20)

Summing eq.(14.19) and eq.(14.20) we get

2∑k

ξkv2k +

∑k

ξk(u2k − v2k)(A

†kAk +B†

kBk)

− 2∑k

ξkukvk(AkBk +B†kA

†k)

− 1

V[∑kk′

Wkk′ukvkuk′vk′

+∑k,k′

Wkk′(−2)ukvkuk′vk′(A†k′Ak′ +B†

k′Bk′)

+∑k,k′

Wkk′ukvk(u2k′ − v2k′)(Bk′Ak′ + A†

k′B†k′)] (14.21)

Requiring the vanishing of the term AkBk +B†kA

†k one gets

2ξkukvk =1

V

∑k′

Wkk′(u2k − v2k)uk′vk′ (14.22)

so that the total Hamiltonian is

H =∑k

Ek(A†kAk +B†

kBk) + E0 (14.23)

100

with

Ek = ξk(u2k − v2k) +

2

V

∑k′

Wkk′uk′vk′ukvk (14.24)

and

E0 = 2∑k

ξkv2k −

1

V

∑kk′

Wkk′ukvkuk′vk′ (14.25)

The operators A†k, B

†k(Ak, Bk) are the creation (annihilation) operators of

quasi-particles.

The eq.(14.22) can be rewritten as

ξk sin 2θk =1

2V

∑k′

sin 2θk′Wkk′ cos 2θk (14.26)

The eq.(14.24), using eq.(14.26), can be rewritten as

Ek = ξk cos 2θk + ξk(sin 2θk)

2

cos 2θk

=ξk

cos 2θk(14.27)

By eliminating ξk in eq. (14.26) using eq.(14.27), we obtain

Ek cos 2θk sin 2θk =1

2V

∑k′

sin 2θk′Wkk′ cos 2θk (14.28)

or

Ek sin 2θk =1

2V

∑k′

sin 2θk′Wkk′ (14.29)

By defining∆k = Ek sin 2θk (14.30)

we have

∆k =2V∑

k′Wkk′

∆k′

Ek′(14.31)

with

Ek =√ξ2k +∆2

k (14.32)

101

Using the explicit form of Wkk′ ,

∆k =W

2V

∑k′

∆k′

Ek′(14.33)

we see that ∆ does not depend on k

∆ =W

2V

∑k′

∆

Ek′(14.34)

We can now study the gap equation (14.34) which has the gapless trivialsolution ∆ = 0. Looking for a solution with ∆ 6= 0, we obtain

1 =W

2V

∑k′

1√ξ2k′ +∆2

(14.35)

This equation can be studied by going into the continuum

1 =W

2VV

1

(2π)3

∫d3k

1√ξ(k)2 +∆2

=W

2

1

(2π)3

∫d3k

1√ξ(k)2 +∆2

(14.36)where the integral is performed around the Fermi surface |ξ(k)| ≤ ~ωD.Notice that there is no solution for W < 0, case corresponding to a repulsiveforce.

14.2 Study of the gap equation

Let us now study the gap equation

1 =W

2

1

(2π)3

∫|ξ(k)|≤~ωD

dΩk2dk1√

(ε(k)− εF )2 +∆2

=W

2

1

(2π)3

∫|ξ(k)|≤~ωD

dΩk2dk

dε

dε√(ε− εF )2 +∆2

∼ W

4ρF

∫|ξ(k)|≤~ωD

dε√(ε− εF )2 +∆2

=W

4ρF

∫ +~ωD

−~ωD

dξ√ξ2 +∆2

=W

2ρFarcsinh

~ωD∆

(14.37)

102

where we have introduced the density of states at the Fermi surface

ρF = 24π

(2π)3k2dk

dε|kF

=1

π2

k2

dεdk

|kF

=1

π2

kFm

~2|kF

(14.38)

where we have introduced the Fermi momentum ~kF .

Inverting eq.(14.37), we get the gap energy. If WρF/2 << 1,

∆ = 2~ωD exp (− 2

WρF) = 2~ωD exp (− 2π2~2

WmkF) (14.39)

For typical metals WρF ∼ 0.3− 0.6, see p.448 of ref. [10]. Then, considering~ωD ∼ 100 K and WρF ∼ 0.6, we get ∆ ∼ 4 K.

In conclusion the energy of the first excitation is given by

Ek =

√(~2k2

2m− εF )2 +∆2 (14.40)

The spectrum has a gap, meaning that one cannot create excitations witharbitrary small energy. The magnitude of this gap is ∆. The quasiparticlesare mixture of electrons and holes (see eq.(14.11)). Furthermore since thequasiparticles have spin 1/2 the quasiparticles must appear in pairs, so theminimim energy is 2∆.

14.3 Finite temperature

Let us now compute how the gap ∆ depends on the temperature. Startingagain from the gap equation, recall that

∆k =1

V

∑k′

Wkk′uk′vk′

103

=1

V

∑k′

Wkk′ < BCS|c−k′,↓ck′,↑|BCS >

=1

V

∑k′

Wkk′uk′vk′ < BCS|[1− (A†k′Ak′ +B†

k′Bk′)]|BCS >

(14.41)

At T = 0, since there is no quasi particles, we recover eq.(14.33) and theformula for ∆ ≡ ∆(T = 0). However this method can be extended at finitetemperature T . Taking the average over a statistical ensemble at temperatureT we have

∆k =1

V

∑k′

Wkk′uk′vk′ < [1−(A†kAk+B

†kBk)] >=

1

V

∑k′

Wkk′uk′vk′(1−2f(Ek′))

(14.42)where f(Ek) is the probability to have an excitations with energy Ek attemperature T :

f(Ek) =1

1 + eβEk(14.43)

Therefore the gap equation at finite temperature becomes

∆k =1

2V

∑k′

Wkk′∆k′

Ek′(1− 2f(Ek′)) (14.44)

Using the explicit expression for W one gets

1 =W

2V

∑k′

1

Ek′(1− 2f(Ek′)) (14.45)

or

−1 +W

2V

∑k′

1

Ek′=W

V

∑k′

f(Ek′)

Ek′(14.46)

Passing to the continuum

−1 +1

2V

V

(2π)3W

∫d3k

1√ξ2(k) + ∆(T )2

=1

V

V

(2π)3W

∫d3kf(E(k))

1

E(k)

(14.47)

104

Let us first compute the l.h. side. Proceeding as before, we get

l.h.side = −1 +1

2WρFasinh

~ωD∆(T )

= −1 +1

2WρF ln

2~ωD∆(T )

=1

2WρF (ln

2~ωD∆(T )

− 2

WρF)

=1

2WρF (ln

2~ωD∆(T )

+ ln∆(0)

2~ωD)

=1

2WρF ln

∆(0)

∆(T )(14.48)

Summing up we have

1

2WρF ln

∆(0)

∆(T )=

1

2WρF

∫ ~ωD

−~ωD

dξ1√

ξ2 + (∆(T ))21

eβ√ξ2+(∆(T ))2 + 1

∼ 1

2WρF

∫ ∞

−∞dx

1√x2 + u2

1

e√x2+u2 + 1

(14.49)

with u = β∆(T ), x = βξ. The integral has been extended to (−∞,∞)because of its rapid convergence. So:

ln∆(0)

∆(T )= 2

∫ ∞

0

dx1√

x2 + u21

e√x2+u2 + 1

(14.50)

This integral is discussed in [9]. For small ∆,

ln∆(0)

∆(T )∼ ln

πkBT

γ∆(T )+

7ζ(3)

8π2

(∆(T ))2

(kBT )2(14.51)

where γ/π ∼ 0.57 and ζ(3) ∼ 1.2. For ∆ = 0 we get the critical temperature:

kBTc =γ

π∆(0) ∼ 0.57∆(0) (14.52)

Using eq. (14.51), expanding for small T −Tc, one gets that by increasingthe temperature the gap becomes smaller and vanishes at TC as

∆(T ) =

√8π

7ζ(3)kBTc(1−

T

Tc)12 ∼ 3.07kBTc(1−

T

Tc)12 (14.53)

105

Tc(K) ~ωD/kB (K) TF × 104 (K) Wρ/2 ∆(T = 0)/kBTc

BCS 1.76Cd 0.56 164 8.7 0.18 1.60 ± 0.05Al 1.2 375 13.6 0.18 1.68 ± 0.05Sn 3.75 195 11.8 0.25 1.73± 0.05Pb 7.22 96 11.0 0.39 2.15± 0.02

Table 1: Some superconductor properties (From [10, 12, ?])

As shown in Table 14.3, the prediction of BCS theory ∆(T = 0) ∼1.76kBTc is quite well satisfied.

Let us finally show that E0 < 0, so that |BCS > is the real ground state.Using (14.27) we can write

ξk = Ek cos 2θk (14.54)

and using (14.29) we can rewrite E0:

E0 = 2∑k

ξkv2k −

1

V

∑kk′

Wkk′ukvkuk′vk′

= 2∑k

Ek cos 2θk sin2 θk −

1

2

∑k

Ek sin2 2θk

= 2∑k

Ek(cos 2θk sin2 θk − sin2 θk cos

2 θk)

= −2∑k

Ek sin4 θk (14.55)

14.4 The BCS ground state

Let us now study the BCS ground state. It is based on the idea that electronsform Cooper pairs. The BCS vacuum is given by

|BCS >= Πk(uk + vkc†k↑c

†−k↓)|0 > (14.56)

Thus it is a superposition of states of Cooper pairs.

106

This state is normalized:

1 =< BCS|BCS > (14.57)

and< BCS|c†k↑c

†−k↓|BCS >= vkuk (14.58)

Proof:

< BCS|BCS > = < 0|Πk(uk + vkc−k↓ck↑)Πk′(uk′ + vk′c†k′↑c†−k′↓)|0 >

= < 0|ΠkukΠk′uk′ +Πkvkc−k↓ck↑Πk′vk′c†k′↑c†−k′↓|0 >

= Πku2k+ < 0|Πk v

2kc−k↓ck↑Πk′vk′c†k′↑c

†−k′↓δk,k′|0 >

= Πk(u2k + v2k) = 1 (14.59)

where use has been made of anticommutation relations:

< 0|c−k↓ck↑c†k↑c

†−k↓|0 >=< 0|c−k↓(1−c†k↑ck↑)c

†−k↓|0 >=< 0|c−k↓c

†−k↓−c−k↓c

†k↑ck↑c

†−k↑|0 >= 1

(14.60)Furthermore

< BCS|c†k↓c†−k↑|BCS > = < 0|Πk′(uk′ + vk′c−k′↓ck′↑)c

†k↑c

†−k↓

Πk′′(uk′′ + vk′′c†k′′↑c†−k′′↓)|0 >

= vkuk (14.61)

The same result can be obtained by the observation that the ground statemust satisfy

Ak|BCS >= 0, Bk|BCS >= 0, ∀k (14.62)

and therefore|BCS >∼ ΠkAkBk|0 > (14.63)

107

A Fourier transform of the Heaviside distri-

butions 1, δ and θ.

In this section we review the definition of Fourier transform of a temperatedistribution [20]. For any distribution temperate T , defined by the linear andcontinuous functional

T : ϕ→ (T, ϕ) (A.1)

where ϕ belong to the Schwartz space, the Fourier transform FT is definedby

(FT, ϕ) = (T, Fϕ) (A.2)

Fourier transform of the distribution 1. For every test function ϕwe have

(F1, ϕ) = (1, Fϕ) =

∫Fϕ(p)dp = (2π)n/2ϕ(0) = ((2π)n/2δ, ϕ) ∀ϕ ∈ S

orF1 = (2π)n/2δ (A.3)

Fourier transform of the distribution δ.

(Fδ, ϕ) = (δ, Fϕ) = Fϕ(0) =1

(2π)n/2

∫ϕ(x)dx = (

1

(2π)n/2, ϕ) (A.4)

therefore

Fδ =1

(2π)n/2(A.5)

Fourier transform of the distribution θ. Let us compute the Fouriertransform of the Heaviside distribution θ:

Fθ(p) =1√2π

∫ +∞

0

θ(x)e−ipxdx =1

i√2π

1

p− i0(A.6)

108

Using the definition of Fourier transform of a temperate distribution, wehave

(Fθ, ϕ) = (θ, Fϕ) =

∫ +∞

0

(Fϕ)(p)dp

= limε→0+

∫ +∞

0

e−εp(Fϕ)(p)dp

= limε→0+

∫ +∞

0

e−εp1√2π

∫ +∞

−∞ϕ(x)e−ixpdxdp

= limε→0+

1√2π

∫ +∞

−∞ϕ(x)

∫ +∞

0

e−(ε+ix)pdpdx

= limε→0+

1√2π

∫ +∞

−∞ϕ(x)

1

ε+ ixdx

=1

i√2π

limε→0+

∫ +∞

−∞

1

x− iεϕ(x)dx (A.7)

where use has been made of the Fubini theorem.

We have also

Fθ =1

i√2π

1

p− i0= − i√

2πPv

1

p+

√π

2δ(p). (A.8)

where we have made use of

1

p− i0= Pv

1

p+ iπδ(p) (A.9)

B The distribution 1p−i0

Let us considered the tempered distribution log(x+ iy) We have

d

dxlog(x+ iy) =

1

x+ iy(B.1)

Therefore

limy→0+

1

x+ iy= lim

y→0+

d

dxlog(x+ iy) (B.2)

109

On the other hand we have

limy→0+

log(x+ iy) = limy→0+

[log |x+ iy|+ iArg(x+ iy)]

= log |x|+ iπθ(−x) (B.3)

By considering the corresponding distributions se have

limy→0+

∫log(x+ iy)ϕ(x)dx =

∫[log |x|+ iπθ(−x)]ϕ(x)dx (B.4)

Therefore using (B.3) we obtain

1

x+ i0= lim

y→0+

1

x+ iy= lim

y→0+

d

dxlog(x+ iy)

=d

dxlimy→0+

log(x+ iy) =d

dx[log |x|+ iπθ(−x)]

= Pv1

x− iπδ(x) (B.5)

In analogous way we have

1

x− i0= Pv

1

x+ iπδ(x) (B.6)

C Coherent states

Aim of this Appendix is the definition and the properties of the coherentstates. The coherent states |c > are defined as eigenvectors of the annihilationoperator of the harmonic oscillator a. Their explicit from is given by

|c >= A1/2

∞∑n=0

cn√n!|n > (C.1)

where

|n >= (a†)n√n!

|0 > (C.2)

andA = exp (−|c|2) (C.3)

110

c is a complex number since the operator a is not hermitian. These statessatisfy the following properties:

< c|c >= 1 (C.4)

|c >= A1/2 exp (ca†)|0 > (C.5)

a|c >= c|c > (C.6)

< c|N |c >=< c|a†a|c >= |c|2 (C.7)

In fact we have

< c|c > = A∑n

∑n′

< n|n′ >1√

n!√n′!

(c∗)ncn

= exp (−|c|2) exp (|c|2) = 1 (C.8)

|c >= A1/2

∞∑n=0

cn√n!

(a†)n√n!

|0 >= A1/2 exp (ca†)|0 > (C.9)

a|c >=∞∑n=1

cn√(n− 1)!

|n− 1 >= c|c > (C.10)

Finally eq.(C.7) derives directly from eq.(C.6).

As an application let us build the coherent states for the electromag-netic field using the destruction operator of a photon with momentum k andpolarization α

|cαk >= A1/2

∞∑n=0

(cαk)n

√n!

|n > (C.11)

with

|n >= (aαk)n

√n!

|0 > (C.12)

We can then compute the expectation value of the electric field E(t,x) =−A(t,x) which turns out to be equal to

< cαk|E(t,x)|cαk >= εαk

√2ωkV

|cαk| sin (k · x− ωkt+ δ) (C.13)

111

wherec = |c| exp (iδ) (C.14)

If we instead consider the expectation value of the electric field E(t,x) on astate with a definite number of photons |Nα1

k1>, the result is zero

< Nα1k1|E(t,x)|Nα1

k1>= 0 (C.15)

In conclusion the coherent state is the quantum state which is closer to aclassical wave.

D Yukawa potential

Let us consider the static equation for the Klein-Gordon field in presence ofa point-like source

(−∇2 +m2)E(x) = δ3(x) (D.1)

By Fourier transforming, we obtain

(q2 +m2)FE(q) = (2π)−3/2 (D.2)

where FE(q) denotes the Fourier transform of E. Therefore

FE(q) =1

(2π)3/21

q2 +m2(D.3)

where q = |q|, and

E(x) =1

(2π)3/2

∫d3qFE(q) exp (iq · x) (D.4)

Therefore

E(x) =1

(2π)3

∫d3q

1

q2 +m2exp (iq · x)

=1

(2π)3

∫ ∞

0

dqq2

q2 +m2

∫ 1

−1

d cos θ

∫ 2π

0

dφ exp (iqx cos θ)

=1

2π2

∫ ∞

0

dqq2

q2 +m2

sin qx

qx

=1

4π2

∫ ∞

−∞dq

q

q2 +m2

sin qx

x

(D.5)

112

where we recall x = |x|. The integral can be compute in the complex planeby using sin qx = (2i)−1(eiqx − e−iqx) and closing the contour above for thefirst exponential or below for the second. By using the residue theorem weget

E(x) =e−m|x|

4π|x|(D.6)

In conclusion we get a screened Coulomb potential with a range of orderm−1. For m = 0 we recover, apart the sign, the Coulomb potential.

E Dirac equation solutions and their proper-

ties

E.1 Spinors

Solutions of the Dirac equations can be written, using the solutions in therest frame of the electrons, as

ur(p) =p+m√

2m(E +m)ur(0), vr(p) =

−p+m√2m(E +m)

vr(0), r = 1, 2 (E.1)

i) the solutions are normalized

ur(p)us(p) = δrs, vr(p)vs(p) = −δrs (E.2)

In fact we have

ur(p)us(p) =1

2m(m+ E)u†(0)r(p

† +m)γ0(p+m)us(0)

=1

2m(m+ E)u(0)γ0(p† +m)γ0(p+m)us(0)

=1

2m(m+ E)u(0)(p+m)2us(0)

=1

(m+ E)ur(0)(p+m)us(0)

= ur(0)us(0)

= δrs (E.3)

113

where we have usedur(0)γ

ius(0) = 0 (E.4)

We have in fact

ur(0)γius(0) = ur(0)

†γius(0) = u†r(0)γiγ0us(0) = −ur(0)γius(0) = 0 (E.5)

In similar way one can prove the normalization for the v.

ii) By introducing the two-component non relativistic spinors

χ1 = η1 =

(10

), χ2 = η2 =

(01

)(E.6)

the solutions can be written using the Dirac-Pauli representation of the Diracmatrices as

ur(p) = A

(χr

Bp · σχr

), vr(p) = A

(Bp · σηr

ηr

), (E.7)

with

A =

(E +m

2m

)1/2

, B =1

E +m(E.8)

The explicit form is

u1(p) = A

10Bp3

B(p1 + ip2)

, u2(p) = A

01

B(p1 − ip2)−Bp3

(E.9)

v1(p) = A

B(p1 − ip2)

−Bp310

, v2(p) = A

Bp3

B(p1 + ip2)01

(E.10)

In fact we have

ur(p) =p+m√

2m(E +m)ur(0)

=1√

2m(E +m)

(E +m −p · σp · σ −E +m

)(χr0

)=

(Aχr

ABp · σχr

)(E.11)

114

iv) The spinors satisfy

u†r(p)us(p) = δrsE

m= v†r(p)vs(p) (E.12)

In fact we have

u†r(p)us(p) =1

2m(m+ E)u†r(0)(p

† +m)(p+m)us(0)

=1

2m(m+ E)ur(0)(p+m)γ0(p+m)us(0)

=1

m(m+ E)ur(0)E(m+ p)us(0)

=1

m(m+ E)ur(0)E(m+ E)us(0)

=E

mδrs (E.13)

where we have used

(p+m)γ0(p+m) = (p+m)(γ0γµpµ + γ0m)

= (p+m)(2g0µpµ − pγ0 + γ0m)

= (p+m)(2E − pγ0 + γ0m)

= 2E(p+m) + (p+m)(−p+m)γ0

= 2E(p+m) (E.14)

Similarly for v.

v) The spinors satisfy also

u†(p)v(−p) = 0 = v†(p)u(−p) (E.15)

andu(p)v(p) = 0 (E.16)

In fact

u†(p)v(−p) = u†(0)(p† +m)(−Eγ0 − pkγk +m)v(0)

= u(0)γ0(p† +m)(−Eγ0 − pkγk +m)v(0)

= u(0)(p+m)γ0(−Eγ0 − pkγk +m)v(0)

= u(0)(p+m)(−Eγ0 + pkγk +m)γ0v(0)

= −u(0)(E2 − p2 −m2)γ0v(0)

= 0 (E.17)

115

u(p)v(p) = u†(0)(p† +m)γ0(−p+m)v(0)

= u†(0)γ0(p+m)(−p+m)v(0)

= −u(0)(E2 − p2 −m2)v(0)

= 0 (E.18)

Solutions of the Dirac equation can also be built using the helicity oper-ator defined as

h(p) =σ · p|p|

(E.19)

where σ is the 4 by 4 matrix

σ =

(σ 00 σ

)(E.20)

This operator, which commutes with the Hamiltonian H = α · p + βm,corresponds to the projection of twice the spin of the particle in the directionof motion. Since

h(p)2 = 1 (E.21)

the eigenvalues of the elicity operator are ±1. Solutions of the Dirac equationin terms of helicity spinors can be found in [5].

E.2 Projection operators

The energy projection operators are

Λ±(p) =±p+m

2m(E.22)

They satisfy

(Λ±(p))2 = Λ±(p), Λ+(p) + Λ−(p) = 1, Λ+(p)Λ−(p) = 0 (E.23)

They project out positive and negative energy states

Λ+(p)ur(p) = ur(p), Λ−(p)vr(p) = vr(p) (E.24)

116

One can easily show that

Λ+αβ(p) =

∑r

urα(p)urβ(p), (E.25)

In fact, in the rest frame, we have

∑r

ur(0)ur(0) =

1000

( 1 0 0 0 ) +

0100

( 1 0 0 0 )

=

1 0 0 00 1 0 00 0 0 00 0 0 0

=

1 + γ0

2(E.26)

Therefore

Λ+ =∑r

ur(p)ur(p) =1

2m(m+ E)(p+m)

∑r

ur(0)u†r(0)(p

† +m)γ0

=1

2m(m+ E)(p+m)

∑r

ur(0)u†r(0)γ

0γ0(p† +m)γ0

=1

2m(m+ E)(p+m)

1 + γ0

2(p+m)

=1

4m(m+ E)[(p+m)2 + (p+m)γ0(p+m)]

=1

2m(m+ E)(m2 + pm+ E(p+m))

=p+m

2m(E.27)

where we have used (E.14). In analogous way

Λ−αβ(p) = −

∑r

vrα(p)vrβ(p) =−p+m

2m(E.28)

117

One can build also spin projectors. In the rest frame they are simply

P±S =

1± σ122

=1

2

(1± σ3 0

0 1± σ3

)(E.29)

with σ12 given by eq.(10.47). They project spin ±1/2 solutions in the 3d di-rection. In particular P+

S projects u1(0), v1(0) while P−S projects u2(0), v2(0).

In a general frame one prefers to consider

P±S =

1± γ5n

2(E.30)

where nµ is a space like four vector orthogonal to pµ

n2 = −1, nµpµ = 0 (E.31)

Now, in the rest frame

PS =1± σ12γ0

2=

1

2

(1± σ3 0

0 1∓ σ3

)(E.32)

and therefore P+S projects u1(0), v2(0) while P

−S projects u2(0), v1(0) .

In the rest frame one has n0 = 0 and therefore one can always choosen = (0, 0, 1).

F Calculation of µ→ eνeνµ decay squared am-

plitude

Aim of this Appendix is the calculation of the sum over the final spins andthe average of the initial spin of the squared amplitude

1

2

∑ri,rf

|Mfi|2 (F.1)

where

Mfi = urνe (pe)γλ(1− γ5)vrνe (pνe)urνµ (pνµ)γλ(1− γ5)urµ(pµ) (F.2)

118

We have

1

2

∑ri,rf

|Mfi|2 =1

2

∑ri,rf

ure(pe)γλ(1− γ5)vrνe (pνe)urνµ (pνµ)γλ(1− γ5)urµ(pµ)

u†rµ(pµ)(1− γ5)γ†σγ0urνµ (pνµ)vrνe (pνe)

†(1− γ5)γσ†γ0ure(pe)

=1

2

∑ri,rf

vrνe (pνe)γ0(1− γ5)γσ†γ0ure(pe)ure(pe)γλ(1− γ5)vrνe (pνe)

urνµ (pνµ)γλ(1− γ5)urµ(pµ)urµ(pµ)γ0(1− γ5)γ

†σγ0urνµ (pνµ)

=1

2Tr[(−Λ−(pνe))γ0(1− γ5)γ

σ†γ0Λ+(pe)γλ(1− γ5)]

Tr[Λ+(pνµ)γλ(1− γ5)Λ

+(pµ)γ0(1− γ5)γ†σγ0

=1

2Tr[(−Λ−(pνe))γ

σ(1− γ5)Λ+(pe)γλ(1− γ5)]

Tr[Λ+(pνµ)γλ(1− γ5)Λ

+(pµ)γσ(1− γ5)]

=1

2

1

16mµmνemνµme

Tr[pνeγσ(1− γ5)peγλ(1− γ5)]

Tr[pνµγλ(1− γ5)(pµ +mµ)γσ(1− γ5)] (F.3)

where in the projection operators Λ we have neglected the masses of thelectron and neutrinos with respect to the µ mass.

Now

Tr[pνeγσ(1− γ5)peγλ(1− γ5)] = 2Tr[pνeγσpeγ

λ]− 2Tr[pνeγσγ5peγλ] (F.4)

Therefore we need

Tr[γαγσγδγλ] = 4(gασgδλ − gαδgσλ + gαλgσδ) (F.5)

andTr[γαγσγ5γδγλ] = −4iεασδλ (F.6)

Therefore

Tr[pνeγσ(1− γ5)peγλ(1− γ5)] = 2Tr[pνeγσpeγ

λ]− 2Tr[pνeγσγ5peγλ]

= 8pανepδeχ

λασδ (F.7)

whereχασδλ = gασgδλ − gαδgσλ + gαλgσδ − iεασδλ (F.8)

119

Furthermore we have

Tr[pνµγλ(1− γ5)(pµ +mµ)γσ(1− γ5)] = Tr[pνµγλ(1− γ5)pµγ

σ(1− γ5)]

+ Tr[pνµγλ(1− γ5)mµγσ(1− γ5)]

= Tr[pνµγλ(1− γ5)pµγσ(1− γ5)]

+Tr[pνµγλmµγσ]− Tr[pνµγλγ5mµγ

σ]

−Tr[pνµγλmµγσγ5] + Tr[pνµγλγ5mµγ

σγ5]

= Tr[pνµγλ(1− γ5)pµγσ(1− γ5)]

= pτνµpρµχ

στλρ (F.9)

where we have usedTr[γµγνγρ] = 0 (F.10)

and the properties of the γ5 matrix. Using

χ λασδ χ

στλρ = 4gαρgστ (F.11)

and substituting in (F.3), we obtain

1

2

∑ri,rf

|Mfi|2 =1

2

1

16mµmνemνµme

Tr[pνeγσ(1− γ5)peγλ(1− γ5)]

Tr[pνµγλ(1− γ5)(pµ +mµ)γσ(1− γ5)]

= 81

mνµmνemµme

(pµ · pνe)(pνµ · pe) (F.12)

G Bose Einstein and Fermi Dirac statistics

Let us now review the quantum statistics. We know from Quantum Mechan-ics that there are two types of particles, bosons and fermions. Single statescan be occupied by any number of bosons while for fermions a single statecan be occupied at most by one fermion.

Since atoms are composed of spin 1/2 particles (neutrons, protons andelectrons) there are atoms which are bosons (H1, He4) and atoms which arefermions (H2, He3). Let us now compute the gran partition function forfree bosons and fermions.Thermodynamic quantities are derived by the gran

120

partition function Z, since Z is connected with the thermodynamic potentialvia

Ω = −kT logZ (G.1)

and the average number of particles and the gas pressure are given by:

N = −∂Ω∂µ

, p = −∂Ω∂V

(G.2)

Let H be the Hamiltonian for N free particles

H =N∑i=1

p2i

2m(G.3)

Let us suppose that for every momentum p there are np particles withsuch momentum. Since we are working in a box, p = 2π/Lm with m =(mx,my,mz) integers. Therefore we have

E =∑p

npεp ≡ E(np) N =∑p

np (G.4)

The gran partition function is given by

Z(µ, V, T ) =∑N

∑np

g(np) exp(−βE(np) + βµN) (G.5)

with g(np) = 1 since all particles are identical. Fer fermions np = 0, 1 whilefor bosons np = 0, 1, 2, · · ·. So we have

Z(µ, V, T ) =∞∑N=0

∑np

exp(−βE(np) + βµN)

=∞∑N=0

∑np

exp[−β∑p

(npεp − µnp)]

=∞∑N=0

∑np

Πp [exp(β(µ− εp))]np

=∑n0

∑n1

[exp(β(µ− ε0))]n0 [exp(β(µ− ε1))]

n1 · · ·

= Πp

∑np

[exp(β(µ− εp))]np

= ΠpZp (G.6)

121

Let us now consider a gas of fermions, then

ZFp =

∑np=0,1

[exp(β(µ− εp))]np = 1 + exp[β(µ− εp)] (G.7)

For a boson gas we have

ZBp =

∑np=0,1,2,...

[exp(β(µ− εp))]np =

1

1− exp[β(µ− εp)](G.8)

Note that in the boson case the series converges only if

exp β(µ− εp) < 1 (G.9)

Therefore if the ground level is for ε0 = 0 the chemical potential must benegative. Finally we can calculate the thermodynamic potential:

ΩF = −kT∑p

ln[1 + exp(β(µ− εp))]

ΩB = kT∑p

ln[1− exp(β(µ− εp))] (G.10)

Given the energy εp we can calculate all the thermodynamic quantities. Letus first compute the average number:

NF = −∂ΩF

∂µ≡

∑p

< np >=∑p

1

exp(β(εp − µ)) + 1

NB = −∂ΩB

∂µ≡

∑p

< np >=∑p

1

exp(β(εp − µ))− 1(G.11)

At low temperature bosons tend to accumulate in the ground state (p = 0),only thermal fluctuations can invert the process. In fact NB increases whenεp − µ→ 0.

The classical limit, the Boltzmann distribution, is obtained for exp(β(εp−µ)) >> 1 or exp(β(µ − εp)) << 1. This corresponds to exp(µ/kT ) <<exp(εp/kT ) or large T and µ/kT → −∞.

Let us now compute the fermion partition function ΩF for a free particlegas by going in the continuum (V → ∞):

ΩF = −kT 4πV

h3

∫ ∞

0

p2dp ln[1 + exp((β(µ− p2

2m))] (G.12)

122

We can now derive average pressure and number of fermions as

p = −∂ΩF

∂V= kT

4π

h3

∫ ∞

0

p2dp ln[1 + exp(β(µ− p2

2m))] (G.13)

N =4πV

h3

∫ ∞

0

p2dp1

1 + exp(β(−µ+ p2

2m))

(G.14)

All the results can be expressed in terms of the functions

f3/2(z) = zd

dzf5/2(z) =

∞∑l=1

(−)l+1 zl

l3/2

f5/2(z) =4√π

∫ ∞

0

dxx2 ln(1 + z exp(−x2)) =∞∑l=1

(−)l+1 zl

l5/2(G.15)

where z = exp (βµ):

p

kT=

1

λ3f5/2(z)

N

V=

1

λ3f3/2(z) (G.16)

where

λ =

√2π~2mkBT

(G.17)

For a Bose gas the partition function is singular when p = 0 and µ → 0or z → 1. Therefore it is convenient to separate in the sum the term withp = 0 before passing to the continuum.

We get

p = −kT 4π

h3

∫ ∞

0

p2dp ln[1− exp(β(µ− p2

2m))]− kT

Vln[1− exp(βµ))] (G.18)

N = V4π

h3

∫ ∞

0

p2dp1

−1 + exp(β(−µ+ p2

2m))

+N0 (G.19)

where

N0 =exp(βµ)

1− exp(βµ)≡ z

1− z(G.20)

123

N0 denotes the number of particles in the p = 0 state.

The results can now be written in terms of the functions

g3/2(z) = zd

dzg5/2(z) =

∞∑l=1

zl

l3/2

g5/2(z) = − 4√π

∫ ∞

0

dxx2 ln(1− z exp(−x2)) =∞∑l=1

zl

l5/2(G.21)

We have

p

kT=

1

λ3g5/2(z)−

1

Vln(1− z)

N

V=

1

λ3g3/2(z) +

N0

V(G.22)

G.1 A gas of free fermions

Let us now rewrite the equations (G.16) for the pressure and concentrationfor a gas of fermions:

p

kT=

1

λ3f5/2(z)

N

V=

1

λ3f3/2(z) (G.23)

The equation of state is obtained by eliminating z from the two equations.Let us start with the equation

λ3

v= f3/2(z) (G.24)

Therefore it is convenient to study the function f3/2 as a function of z. f3/2is a function monotone in z. For small z

f3/2(z) = z − z2

23/2+

z3

33/2− z4

43/2+ · · · (G.25)

For large z (Huang p.246)

f3/2(z) =4

3√π[(ln z)3/2 +

π2

8

1

(ln z)1/2] +O(1/z) (G.26)

124

Therefore for every positive value of λ a solution for z exists.

Low density and high temperature, λ3/v << 1

In this case the thermal length λ ∼ ~/p ∼ ~/√mKT is much smaller than

the average distance among the particles v1/3, therefore quantum effects arenegligible. From

λ3

v= z − z2

23/2+ . . . (G.27)

one gets

z =λ3

v+

1

23/2(λ3

v)2 + . . .) (G.28)

and the equation of state becomes

pV

kTN=

v

λ3(z − z2

25/2+ . . . = 1 +

1

25/2λ3

v+ . . . (G.29)

Therefore one obtains quantum corrections to the classic case. Recalling thevirial expansion

p

kT=N

V(1 +B(T )

N

V+ C(T )

(N

V

)2

+ . . .) (G.30)

one can identify the virial coefficients.

High density and low temperature, λ3/v >> 1

In this case the thermal distance is much larger than the average distanceso the quantum effects become relevant. The leading term is now

λ3

v=

4

3√π(ln z)3/2 (G.31)

Therefore we getz = exp (βεF ) (G.32)

where the chemical potential εF is called Fermi energy

εF =~2

2m

[6π2

v

]2/3(G.33)

125

Let us now consider < np >, defined in eqs.(G.11)

< np >=1

eβ(εp−εF ) + 1(G.34)

When T → 0, β → +∞ we have

< np >T=0= 1 (G.35)

for εp < εF and< np >T=0= 0 (G.36)

for εp > εF .

Therefore at zero temperature the fermions occupy all the lowest levels upto εF . Because of the Pauli principle they cannot occupy all the ground stateand therefore they fill all the states up to the highest energy εF . Such a stateis called a degenerate Fermi gas. In the momentum space the particles fill asphere of radius pF , the Fermi surface. One defines also a Fermi temperatureor degeneracy temperature TF such that

kTF = εF (G.37)

Finally we can compute the internal energy

U =∑p

εpnp =V

h34π

2m

∫ ∞

0

dpp4 < np > (G.38)

By part integration we get

U =V

4π2m~3

∫ ∞

0

dpp5

5

(− ∂

∂pnp

)=

βV

20π2m2~3

∫ ∞

0

dpp6eβεp−βµ

(eβεp−βµ + 1)2

(G.39)

The integrand has a peak at p = pF . The asymptotic behavior of theintegral is (see [?])

U =3

5NεF

[1 +

5

12π2(

kT

εF)2]

(G.40)

From the internal energy one can derive the specific heat

CV = NkBπ2

2

kT

εF(G.41)

126

which goes to zero when T → 0 (Third law of thermodynamics) and thepressure

p =2

3

U

V=

2

5

εFv

[1 +

5

12π2(

kT

εF)2]

(G.42)

Notice that even at T = 0 as a consequence of Pauli principle the gas hasa non vanishing pressure. This pressure is responsible for the gravitationalstability of white dwarfs and neutron stars. A white dwarf can be thought as agas of ionized helium and electrons. The gravitational stability is guaranteedby the degenerate electron pressure. In the case of neutron stars the stabilityis guaranteed by the pressure of the degenerate gas of neutrons.

Note Alternative way to compute εF .

An alternative way of computing εF is to fill all the states in the momen-tum space up to pF :

N =4πV

h3

∫ pF

0

p2dp =4πV

3h3p3F =

V

6π2~3p3F (G.43)

or

pF = (N

V)1/3(6π2)1/3~ (G.44)

εF =1

2m(N

V)2/332/3h2 =

~2

2m

[6π2

v

]2/3(G.45)

G.2 A gas of free bosons

Let us now study with some detail the Bose case. The function g3/2 for zsmall can be studied as a series

g3/2(z) = z +1

23/2z2 +

1

33/2z3 + · · · (G.46)

with

g3/2(1) =∞∑l=1

1

l3/2= ζ(

3

2) = 2.612... (G.47)

where ζ is the Riemann function. As we have already noticed for a boson gasµ < 0 then 0 ≤ z ≤ 1 and g3/2 ≤ 1. Rewriting the eq.(G.22) for the averagenumber

λ3N0

V=λ3

v− g3/2(z) (G.48)

127

with

v =V

N(G.49)

we see that N0/V > 0 if temperature and specific volume v are such that

λ3

v> g3/2(1) (G.50)

In fact

0 <λ3

v− g3/2(1) <

λ3

v− g3/2(z) (G.51)

This means that the ground state is occupied by a macroscopic fraction ofbosons. The critical temperature for the Bose condensation is defined by

λ3cv

=1

v

(2π~2

mkTc

)3/2

= g3/2(1), or µ = 0, N0 = 0 (G.52)

or

Tc =1

kB

2π~2/m[vg3/2(1)]2/3

=1

kB

2π~2/m[ζ(3/2)]2/3

(N

V)2/3 (G.53)

At the critical temperature the bosons start to occupy the p = 0 state andif the temperature decreases more and more bosons occupy such a state. ForT < Tc the chemical potential µ remain zero. Inserting the values ρHe4 =0.145g/cm3 ∼ mHe4N/V with mHe4 = 4mp, mp = 4 × 1.67 × 10−27Kg,~ = 1.05510−34 J s, the Boltzmann constant k = 1.38 10−23J/K) we getTc ∼ 3.14 K. This temperature is very close to the critical temperature ofliquid Helium, Tλ = 2.17 K, below which the helium becomes superfluid.

H Fundamental state of the superfluidity the-

ory

Let us now discuss the properties of the new vacuum state |φ0 > of theHilbert space of quantum states of superfluidity. The usual form of quantumfield theory vacuum cannot be used since the fundamental state for a systemof N bosons is given by

|φ0(N) >= |N, 0, · · · 0 > (H.54)

128

that means that all the particles are in the lowest energy state (k = 0).Therefore the annihilation operator a0 does not annihilate the minimum en-ergy state

a0|0 >= 0 (H.55)

buta0|φ0(N) >= N1/2|φ0(N − 1) > (H.56)

ea†0|φ0(N) >= (N + 1)1/2|φ0(N + 1) > (H.57)

To find the minimum energy state, it is necessary to consider first the coher-ent state

|φ0 >= A1/2 exp[√V φ0a

†0]|0 > (H.58)

which satisfiesa0|φ0 >=

√V φ0|φ0 > (H.59)

andak|φ0 >= 0 k 6= 0 (H.60)

n0 =< N(k = 0) >

V=

1

V< φ0|a†0a0|φ0 >=

1

VV φ2

0 (H.61)

In other words the expectation value of N is V φ20. The normalization is given

by

A1/2 = exp[−1

4V φ2

0] (H.62)

Therefore n0 is the boson density in the state k = 0. The vacuum expectationvalue of the field φ(x) on the state |φ0 > is given by

< φ0|φ(x)|φ0 >= φ0 =√n0 =

√< N(k = 0) >

V(H.63)

and it is related to the density of the condensate. The true vacuum state ishowever defined as we have found in Section 13.4 by the condition (13.57)

Ak|φ0 >=

[cosh(

θk2)ak + sinh(

θk2)a†−k

]|φ0 >= 0 (H.64)

The solution is given by

|φ0 >= N exp [−1

2

∑k 6=0

tanh(θk/2)a†ka

†−k]|φ0 > (H.65)

129

This means that the true vacuum state contains pair of bosons with oppositemomenta.

Exercise. Verify the at eq. (H.64) is satisfied by the new vacuum (H.65).

Exercise. Verify that the state |φ0 > corresponds to a lower value of theenergy with respect to |φ0 >.

I Bogoliubov transformation

Let us now derive the Bogoliubov transformation. Let us start considering∑k 6=0

[αka

†kak +

µ

2(aka−k + a†ka

†−k)

](I.1)

where

αk = µ+~2k2

2m(I.2)

Let us considerAk = βkak + γka

†−k (I.3)

with βk, γk ∈ R. Then we get

[Ak, A†k′ ] = [βkak + γka

†−k, βka

†k′ + γka−k′ ] = (β2

k − γ2k)δkk′ (I.4)

In order to get standard commutation relations, let us require

β2k − γ2k = 1 (I.5)

It is convenient to define

βk = cosh

(θk2

), γk = sinh

(θk2

)(I.6)

The inverse transformations are

ak = βkAk − γkA†−k, a†k = βkA

†k − γkA−k (I.7)

In fact

βkAk − γkA†−k = βk(βkak + γka

†−k)− γk(βka

†−k + γkak) = ak (I.8)

130

Substituting in eq.(I.1) one obtains∑k 6=0

[αka

†kak +

µ

2(aka−k + a†ka

†−k)

]=

∑k 6=0

[αk(βkA

†k − γkA−k)(βkAk − γkA

†−k)

+µ

2

((βkAk − γkA

†−k)(βkA−k − γkA

†k)

+(βkA†k − γkA−k)(βkA

†−k − γkAk)

)]=

∑k 6=0

[(β2k + γ2k)αk − 2βkγkµ)A

†kAk +

(−βkγkαk +µ

2(β2

k + γ2k))(AkA−k + A†kA

†−k)

+αkγ2k − βkγkµ] (I.9)

By requiring the vanishing of the coefficient of AkA−k + A†kA

†−k we get

tanh θk =2βkγkβ2k + γ2k

=µ

αk(I.10)

Then the coefficient of A†kAk becomes, using (I.10) and (I.6)

(β2k + γ2k)αk − 2βkγkµ = (β2

k + γ2k)αk −4β2

kγ2kαk

β2k + γ2k

=(β2

k − γ2k)2αk

β2k + γ2k

=αk

β2k + γ2k

=αk

cosh θk= αk

√1− tanh2 θk =

√α2k − µ2

≡ ε(k) (I.11)

with ε(k) given by eq.(13.59). Finally

αkγ2k − βkγkµ = αkγ

2k

γ2k − β2k

β2k + γ2k

= − αkγ2k

β2k + γ2k

= −ε(k) sinh2

(θk2

)(I.12)

where use has been made of eq.(I.11).

References

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131

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133